The Mean Value Theorem: A Comprehensive Introduction

Hello everyone! Today, we’re going to attempt a deep dive into an over looked and underrated calculus theorem , “The Mean Value Theorem” . We come across the Mean Value Theorem in most undergraduate calculus classes, this theorem not only links the concept of derivatives and average slope but also serves as a” workhorse” theorem underpinning many of the techniques and methods we utilize throughout a standard series of calculus courses.

This is the 1st in a series on the three “Value Theorems” of calculus. Many students see calculus as just a series of disconnected terms and complex rules, making it difficult to grasp the subject’s overarching structure, especially under the pressures of grades and tight deadlines. I am taking the time to dive into foundational concepts like the Value Theorems with the hope of helping students see calculus as a cohesive whole rather than isolated fragments.

So with out further delay…

Introduction

For a function f that is continuous on, [a, b] and differentiable on (a, b), the MVT states there exists at least one point c in (a, b) such that:

f'(c) = \frac{f(b) - f(a)}{b - a}Essentially, this means there’s at least one point (c, f(c)) where the instantaneous rate of change equals the average rate of change over that interval.”

Google likes to puts it as.

It’s always referred to as the Mean Value Theorem No one calls it Lagrange Mean Value Theorem. I guess some people call it Lagrange Mean Value Theorem… Don’t call it the Lagrange Mean Value Theorem.

Problem #1

Some Quick Warm Up Problems…and Desmos tool *open in new tab*

For Problem #2 To #4 For f(x) on the interval [a,b], calculate m = \frac{f(b) - f(a)}{b - a} and find the value(s) of c so that f'(c) = m.

f(x) = x^2 \text{ on } [0,2]

f(x) = x^2 - 5x + 8 \text{ on } [1,5]

f(x) = \sin(x) \text{ on } [0,{\pi}]

Problem #5

The Classic Motivating Example: Speed Limit Enforcement

Problem #6

“Now imagine if you will, a scenario with two traffic officers, monitoring traffic, each equipped with a radar gun. Officer A is stationed at mile marker 10, and Officer B is at mile marker 17, 7 miles apart. A car passes Officer A at 3:00 pm, and the radar shows the vehicle traveling at 60 mph. The car then passes Officer B at 3:06 pm, with the radar indicating a speed of 62 mph. The speed limit on this stretch of road is 65 mph.. Now, the officers face a decision on whether to issue a speeding citation to the driver”

Why would the officers consider giving a ticket for speeding when the vehicle’s speed, as measured at both points, did not exceed the posted limit?”

As can be seen by the above example The Mean Value Theorem has a very natural interpretation when f(x) represents the position of an object at time x: f'(x) represents the velocity of the object at the instant x and \frac{f(b) - f(a)}{b - a} represents the average (mean) velocity of the object during the time interval from time a to time b. The Mean Value Theorem says that there is a time c (between a and b) when the instantaneous velocity, f'(c), is equal to the average velocity for the entire trip, \frac{f(b) - f(a)}{b - a}. So it follows,

“If your average velocity during a trip is 30 miles per hour, then at some instant during the trip you were traveling exactly 30 miles per hour.”

One more traditional physics scenario

Problem #7

After going going through the above sample problems the Mean Value Theorem can quickly become one of the things that makes you think, “What’s the point?” because it seems so obvious that it can be hard to understand that this is worth a proof.

Note this educators attitude towards these kinds of sample problems

…. Hmm Looks to me like one can only appreciate the value of the theorem when they apply it to a problem were it isn’t really obvious that it can be helpful…

Hunting for more real world examples!!!(that have nothing to do with Physics !):

These images should make the mathematical hairs stand up on the back of your neck. !!!

and if in fact you thought what I was thinking …

That made me happy. :-)

Lets poke around… *open in new tab*

OK…

Lets take a look at a proof of the Theorem.

A Proof of the Mean Value Theorem

Given:

1. Let f be a function such that:

   • f is continuous on the closed interval [a, b].
   • f is differentiable on the open interval (a, b).

2. The slope m of the secant line between points a and b on the graph of f is given by m = \frac{f(b) - f(a)}{b - a}.

3. The point-slope form of a line is given by the equation y - y_1 = m(x - x_1), where m is the slope of the line and (x_1, y_1) is a point on the line.

4. Rolle’s Theorem states that if a function g is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and g(a) = g(b), then there is at least one point c in (a, b) such that g'(c) = 0.

   

   

Mean Value Theorem Statement:

There exists at least one point c in (a, b) such that:

f'(c) = \frac{f(b) - f(a)}{b - a} = m

Proof:

The equation for the secant line through (a,f(a)) and (b, f(b)) is y = \left( m \right)(x - a) + f(a)

Then construct a function g(x) incorporating f(x) and the secant line connecting (a, f(a)) and (b, f(b)):

g(x) = f(x) - y

g(x) = f(x) - \left((m \right)(x - a) + f(a))

Properties of g(x):

g(a) = g(b), as both evaluate to 0 \begin{align*}g(a) &= f(a) + \left( -\frac{f(b) - f(a)}{b - a} \right)(a - a) - f(a) \\&= f(a) - 0 - f(a) \\&= 0\end{align*}

and

\begin{align*} g(b) &= f(b) + \left( -\frac{f(b) - f(a)}{b - a} \right)(b - a) - f(a) \\ &= f(b) - f(b) + f(a) - f(a) \\ &= 0 \end{align*}

Also note g(x) is continuous on [a, b] and differentiable on (a, b), mirroring the properties of f. ( because both f and y are continuous on [a, b] and differentiable on (a, b))) This means we can apply Rolle’s Theorem to g(x).

Hence by Rolle’s Theorem there must be at least one point c in (a, b) where g'(c) = 0

The derivative of g(x) is computed as:

g'(x) = f'(x) - \frac{f(b) - f(a)}{b - a}

So at the point c where g'(c) = 0, it follows that:

f'(c) - \frac{f(b) - f(a)}{b - a} = 0

And thus, we have:

f'(c) = \frac{f(b) - f(a)}{b - a}

This completes the proof of the Mean Value Theorem.

A visual of the proof is provided below…

again note these similarities…

So maybe this is not such a pointless exorcise after all…

A note on important classroom applications of the Mean Value Theorem.

How do we know that a derivative equal to zero over an interval signifies the function is constant, because of the Mean Value Theorem.

note we can now go backwards from the rule below

So if f'(x) = 0, then f(x)=c .

How do we demonstrate that a positive(negative) derivative indicates a function is increasing(decreasing)? Using the Mean Value Theorem.

In fact the fundamental idea behind Curve Sketching in general (finding critical and inflection points…etc)…based on Mean Value Theorem in some way.

You know Those +C’s that we ( forget too )tack on to the end of our indefinite integrals … Mean Value Theorem.

We will see later on the important connections that the Mean Value Theorem also has with The Fundamental Theorem of Calculus , Taylor Series, and L’Hôpital’s rule.

These other theorems and methods get all the limelight, but in a one way or another, they all are refined versions or repeated applications of the Mean Value Theorem plus special conditions. All of these widely accepted principles, we often apply without revisiting proofs, would lack formal justification without the Mean Value Theorem.

So always remember

Three more insightful examples, the last of which will come in handy when we get to Cauchy’s Mean Value Theorem

Problem #8 & #9

Problem #10

Zoom in for Solution:

Non-Examples of the Mean Value Theorem

Instances where the Mean Value Theorem does not apply, due to the functions not meeting the necessary conditions can prove to be insightful:

1. Discontinuity: If the function is discontinuous on the interval [a, b], it violates one of the key conditions of the theorem. For instance, a function that has a jump or a point of discontinuity within the interval does not satisfy the Mean Value Theorem.

   

2. Non-differentiability: If the function is not differentiable at any point on the interval [a, b], the theorem cannot be applied. This could occur in cases where the function has a sharp corner or a cusp.

   Problem #11

Problem #12 & Problem #13

This next set of problem is insightful.

Problem #14

Interesting!!!! even though one of the functions failed the conditions of the MVT we were still able to find f’ = m !!!!

why? Because…

Note one more example…

Even if a function does not strictly meet the conditions specified by the Mean Value Theorem , it’s still possible, in certain instances, to find a value c in the interval [a,b] where the derivative of the function at c equals the average rate of change over [a,b]. However, finding such a c in these scenarios is not guaranteed.

Problem #15

Extensions and Variations of The Mean Value Theorem:

As we have seen The Mean Value Theorem is a cornerstone of calculus that provides essential insights into the behavior of differentiable functions. Several variations and extensions of the theorem apply to different mathematical contexts and applications. We high highlight a select few simple examples here. This list is by no means extensive.

Mean Value Theorem for Integrals

The Mean Value Theorem for integrals guarantees that for a function ( f ) continuous on ([a, b]), there exists at least one ( c ) in ([a, b]) such that the definite integral of ( f ) over that interval is equal to the function’s value at ( c ) times the length of the interval. Or that the value The average value of the function equals f(c)

They re-branded this poor guy as “Average” … but we all know … he’s

The required standard example, and Desmos Tool *open in new tab*

Problem #16

A Very Useful Application

Problem #17

Mean Value for Integrals Theorem Statement

There exists at least one point c in (a, b) such that:

\int_{a}^{b} f(x) \, dx = f(c)(b - a)

Proof:

Consider F(x) = \int_{a}^{x} f(x) \, dx

Applying the Fundamental Theorem of Calculus, we know that F'(x) = f(x) for all x in (a, b).

and \int_a^b {f(x)dx}= {F(b) - F(a)}

By applying the Mean Value Theorem to F(x) (necessary conditions easily verifed ) over [a, b], there exists a point c in (a, b) such that:

F'(c) = \frac{F(b) - F(a)}{b - a}

Substituting:

f(c) = \frac{\int_{a}^{b} f(x) \, dx}{b - a}

Multiplying both sides by (b - a), we obtain the result:

\int_{a}^{b} f(x) \, dx = f(c)(b - a)

This result completes the proof of the Mean Value Theorem for integrals.

A visual of the proof is provided below…

Cauchy’s Mean Value Theorem

Description

An extension of the classical MVT, Cauchy’s Mean Value Theorem addresses two functions and states that if ( f ) and ( g ) are continuous on ([a, b]) and differentiable on ((a, b)), then there exists a ( c ) in ((a, b)) such that:

(f(b) - f(a)) g'(c) = (g(b) - g(a)) f'(c)

or \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}

A Simple Motivating Example from - https://www.mathwarehouse.com/calculus/derivatives/what-is-cauchys-mvt.php

Let’s say the two boats A and B are traveling for exactly 15 minutes. At the end of that time, the Boat A covered exactly twice the distance as the Boat B. What Cauchy’s extension says is that in order for Boat A to travel exactly twice the distance of Boat B, Boat A’s speed had to be exactly twice the speed of Boat B at least once during the 15 minute interval.

(Think back to the problem with the runners who finished in a tie, how is this similar … how is it different? )

We kinda lose the standard visual when we use this method or at the very lest its messy… but don’t fear …

Problem #18

Applications

This Cauchy’s theorem is used for handling expressions involving rates of change in two related quantities, often used in physics and economics to model systems involving multiple variables.

For example our old friend:

More details on this example can be found at( No calculus though sadly) https://analystprep.com/cfa-level-1-exam/economics/price-marginal-cost-marginal-revenue-economic-profit-and-the-elasticity-of-demand/

Two different Proofs of Cauchy’s Mean Value Theorem can be found -https://www.mathwarehouse.com/calculus/derivatives/what-is-cauchys-mvt.php

A visual of the proof is provided below … note how similar the proof is to that of the classic MVT, note the use of Rolle’s Theorem.

Also!!!!

Here are two examples

Other Cool Mean Value Theorem Facts!

Taylor’s Theorem

Check this out….

So….

This video makes the connection clear

Honorable Mentions:

There is a more detailed & exhaustive list on the Wiki -_https://en.wikipedia.org/wiki/Mean_value_theorem

This whole thing would be incomplete without this…

To find the slope of lines in Math-e-matics
We often use the phrase rise over run.
Divide the change in y over the x’s
Care-fully, remember how it’s done?
CHORUS:
If f ’s continuous and differenti’ble
On a closed in-terval from a to b.
Slope of the secant line equals f prime sometime,
This the Mean Val-ue Theorem guarantees.

Cal-culate the slope, first, of the se-cant
Also called the average rate of change.
Set f prime at c equal to this number,
Then find all the values c can be.
Sing CHORUS
When your points have got the same y value
Zero is the slope you will agree.
Set f prime at c equal to that zero
Max or min’s are possibilities.
Sing CHORUS
When you need to find the average value
When integrating f from a to b.
You take the integral and just di-vide it
By the length of b minus the a.
Sing CHORUS

Context & History of the Mean Value Theorem

Mathematical Context: The Mean Value Theorem follows the following conceptual “chain”: The Axiom of Completeness -> The Extreme Value Theorem -> Rolle’s Theorem

Axiom of Completeness: Every nonempty set of real numbers that is bounded above has a least upper bound, or supremum. This axiom is fundamental to the real number system, ensuring that the reals are “complete” in the sense that there are no “gaps” in the number line.

Extreme Value Theorem: If a function f is continuous on a closed interval [a,b], then f must attain both a maximum and a minimum value, at least once each, on that interval. In other words, continuous functions on closed intervals are guaranteed to have both highest and lowest values.

Rolle’s Theorem: If a function f is continuous on a closed interval [a,b], differentiable on the open interval (a,b) , and f(a)=f(b), then there is at least one point c in (a,b) at which the derivative f′(c) is zero. This theorem can be viewed as a special case of the Mean Value Theorem where the average rate of change on the interval is zero.

The next section outlines the development of the Mean Value Theorem, tracing its historical evolution from early intuitions to rigorous mathematical formulations.

Early Intuitions and Contributions

14th Century: Parameshvara

Parameshvara from the Kerala school of mathematics in India showed early forms of differential calculus. His insights laid foundational ideas that resonate with the concepts later formalized in the Mean Value Theorem. A special case of this theorem for inverse interpolation of the sine was first described by Parameshvara in his commentaries on Govindasvāmi and Bhāskara II, marking an early understanding of the principles underlying the MVT.

17th Century: Bonaventura Cavalieri

In 1635, Cavalieri proposed an early version of the MVT. However, his proof lacked rigor by modern standards and was not accepted, critiqued by Paul Guldin. Despite the rejection, Cavalieri’s attempt marks a significant early effort to grasp the relationships between average and instantaneous rates of change.

Formal Developments and Theoretical Refinements

Michel Rolle (1691)

Rolle provided a crucial breakthrough with the formal statement and proof of Rolle’s Theorem for polynomial functions, a special case of the MVT. This theorem states that if a function equals its endpoint values and is differentiable in between, its derivative must be zero at some point within the interval. His work was seminal although it lacked the full apparatus of calculus, which was still being developed.

Leonhard Euler (1755)

Euler generalized Rolle’s theorem to differentiable functions, extending the application and understanding of these foundational concepts.

Joseph-Louis Lagrange (1797)

Lagrange proved the MVT for analytic functions in his work “Théorie des fonctions analytiques.” This proof, while initially accepted, was later seen as a precursor to more rigorous formulations.

Rigorous Proofs and Modern Formulations

Augustin-Louis Cauchy (1823)

Cauchy’s 1823 proof of the MVT for functions with continuous first derivatives, and this set a new standard for rigor. His work clarified and solidified the theorem’s application and was instrumental in the development of real analysis. He was the first to state and prove the theorem in its modern form, integrating rigorous calculus techniques.

Pierre-Ossian Bonnet (1868)

Bonnet rearranged the proofs concerning MVT, making critical use of Rolle’s Theorem as a lemma in proving MVT. His contributions further refined the understanding and proof structure of the theorem.

Ulisse Dini (1878)

Dini provided the current standard proof of Rolle’s Theorem, which is now foundational for teaching and understanding MVT. His work closed gaps in previous proofs and offered a robust framework that supports modern calculus education.

Conclusion

The development of the Mean Value Theorem is a testament to the collaborative and cumulative nature of mathematics, involving insights from diverse cultures and centuries of refinement. From Parameshvara’s early calculations to Dini’s rigorous proofs, each step in the theorem’s evolution built on the past, showcasing the intricate interconnections of mathematical ideas and the ongoing quest for precision and understanding in mathematical theory.

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Hidden challenge problem

Solutions

Problem #1

For the function f(x) = 5x^2 - 4x + 3 on the interval [1,3], we first calculate f(3) and f(1):

f(3) = 5(3)^2 - 4(3) + 3 = 36

f(1) = 5(1)^2 - 4(1) + 3 = 4

The average rate of change m on the interval [1,3] is:

m = \frac{f(3) - f(1)}{3 - 1} = \frac{36 - 4}{2} = 16

The derivative f'(x) is:

f'(x) = \frac{d}{dx}(5x^2 - 4x + 3) = 10x - 4

To find the value(s) of c such that f'(c) = m, we solve the equation f'(c) = 10c - 4 = 16:

10c - 4 = 16

10c = 20

c = 2

Therefore, the value of c on the interval [1,3] where the derivative of f equals the average rate of change m is c = 2.

Problem #2

For the function f(x) = x^2 on the interval [0,2], calculate m = \frac{f(b)-f(a)}{b-a} and find the values of c so that f'(c) = m.

Solution

• f(2) = 4
• f(0) = 0
• The average rate of change m on the interval [0,2] is 2.
• The derivative f'(x) is 2x.

Setting the derivative equal to m gives us f'(c) = 2c = 2, which yields c = 1.

Problem #3

For the function f(x) = x^2 - 5x + 8 on the interval [1, 5], calculate m and find the value(s) of c so that f'(c) = m.

Solution

• f(5) = 8
• f(1) = 4
• The average rate of change m on the interval [1,5] is 1.
• The derivative f'(x) is 2x - 5.

Setting the derivative equal to m gives us f'(c) = 2c - 5 = 1, which yields c = 3.

Problem #4

For the function f(x) = \sin(x) on the interval [0, \pi], calculate m and find the value(s) of c so that f'(c) = m.

Solution

• f(\pi) = 0
• f(0) = 0
• The average rate of change m on the interval [0, \pi] is 0.
• The derivative f'(x) is \cos(x).

Setting the derivative equal to m gives us f'(c) = \cos(c) = 0. The solution for c within the interval [0, \pi] is c = \frac{\pi}{2}.

Problem #5

Assume that a function f(x) is continuous and differentiable on the interval [5, 15]. Given f(5) = 4 and f'(x) \leq 10. Find the largest possible value that f(15) can take on.

Solution

Since f'(x) \leq 10, the maximum average rate of change of f(x) from x = 5 to x = 15 is 10. Using the Mean Value Theorem:

f(15) - f(5) \leq 10(15 - 5) f(15) - 4 \leq 100 f(15) \leq 104

The largest possible value that f(15) can take on is 104.

Problem #6

Consider a scenario with two police officers, each equipped with a radar gun. Officer A is stationed at mile marker 10, while Officer B is at mile marker 17, making the distance between them 7 miles. A car passes Officer A at 3:00 pm with the radar indicating a speed of 60 mph. Then, the car passes Officer B at 3:06 pm, and the radar shows a speed of 62 mph. The speed limit on this stretch of road is 65 mph.

The question is whether the car deserves a speeding ticket based on the average speed over the distance between the two officers.

The time between the car passing Officer A and Officer B is 6 minutes. This time needs to be converted into hours to calculate the average speed in miles per hour (mph).

\text{Time in hours} = \frac{6 \text{ minutes}}{60 \text{ minutes per hour}} = \frac{1}{10} \text{ hours}

Using the formula for average speed m, we have:

m = \frac{f(t_b) - f(t_a)}{t_b - t_a}

m = \frac{\text{Distance}}{\text{Time}}

Substituting the given values into the formula gives us:

m = \frac{17 - 10}{\frac{1}{10}} = \frac{7}{\frac{1}{10}}= \frac{7 \text{ miles}}{\frac{1}{10} \text{ hours}} = 7 \times 10 = 70 \text{ mph}

Even though the radar readings at both mile markers indicate the vehicle’s speed as being below the speed limit of 65 mph, the calculated average speed over the 7-mile stretch is 70 mph, which exceeds the speed limit.

According to the Mean Value Theorem, at some point between the two radar checks, the vehicle’s instantaneous speed must have been equal to this average speed. Therefore, the vehicle must have been speeding during the journey between the two officers. Thus, the officers would be justified in issuing a speeding ticket based on the calculated average speed.

Problem #7

A ball is dropped from a height of 100 ft. Its position in seconds after it’s dropped is modeled by the function s(t) = -16t^2 + 100.

1. How long does the ball take to hit the ground after it’s dropped?

   To find the time when the ball hits the ground, we solve s(t) = 0 for t:

   0 = -16t^2 + 100

   t^2 = \frac{100}{16}

   t = \sqrt{\frac{100}{16}} = 2.5 \text{ seconds}

2. Find the ball’s average velocity between when it is released and when it hits the ground.

   The average velocity v_{avg} is given by:

   v_{avg} = \frac{s(2.5) - s(0)}{2.5 - 0}

   Since s(0) = 100 and s(2.5) = 0, the average velocity is:

   v_{avg} = \frac{0 - 100}{2.5} = -40 \text{ ft/sec}

3. Then, find the time guaranteed by the Mean Value Theorem when the instantaneous velocity of the ball is equal to the ball’s average velocity.

   The instantaneous velocity v(t) is the derivative of s(t) with respect to t:

   v(t) = s'(t) = -32t

   Setting the instantaneous velocity equal to the average velocity:

   -32t = -40

   Solving for t gives us:

   t = \frac{40}{32} = 1.25 \text{ seconds}

At t = 1.25 seconds, the instantaneous velocity of the ball is equal to its average velocity.

Problem #8

A marathoner ran the 26.2 mi New York City Marathon in 2.2 h. The Mean Value Theorem assures us that there must be at least one point where the marathoner’s instantaneous speed was equal to their average speed. The average speed can be calculated as follows:

\text{Average speed} = \frac{26.2 \text{ miles}}{2.2 \text{ hours}} \approx 11.91 \text{ mph}

Since the average speed is greater than 11 mph, by the Mean Value Theorem, there must be at least two points during the marathon where the marathoner’s speed was exactly 11 mph.

Problem #9

A plane begins its takeoff at 2:00 PM on a 2500 mile flight. After 5.5 hours, the plane arrives at its destination. The average speed is given by:

\text{Average speed} = \frac{2500 \text{ miles}}{5.5 \text{ hours}} \approx 454.55 \text{ mph}

Even though the average speed is greater than 400 mph, the Mean Value Theorem ensures that there must be at least two moments when the plane’s speed is exactly 400 mph as it increases to cruising speed and decreases for landing.

Problem #10

In text…

Problem #11

Given the function

f(x) = \frac{1}{(x - 3)^2},

we are asked to show there is no value c in ( (1,4) ) such that f'(c)(4 - 1) = f(4) - f(1), and explain why this does not contradict the Mean Value Theorem (MVT).

First, we compute the average rate of change of ( f ) on the interval from 1 to 4:

f(4) = \left(4 - 3\right)^{-2} = 1,

f(1) = \left(1 - 3\right)^{-2} = \frac{1}{4},

\text{Average rate of change} = \frac{f(4) - f(1)}{4 - 1} = \frac{1 - \frac{1}{4}}{3} = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}.

Now, let’s consider the derivative ( f’(x) ), which is:

f'(x) = \frac{d}{dx} \left(\frac{1}{(x - 3)^2}\right) = -\frac{2}{(x - 3)^3}.

According to the MVT we have:

-\frac{2}{(c - 3)^3} = \frac{1}{4}.

Solving for c , we obtain:

c - 3 = -\sqrt[3]{\frac{2}{\frac{1}{4}}}

c = 3 - \sqrt[3]{8}

c = 3 - 2

c = 1.

However, c = 1 is not in the open interval (1,4), therefore it does not satisfy the condition required by the MVT.

The absence of such a value c does not contradict the MVT because the theorem requires that the function be continuous on the closed interval [1, 4] and differentiable on the open interval (1, 4). The function f(x) = (x - 3)^{-2} is not continuous on [1, 4] since it is not defined at x = 3, which lies within this interval. Therefore, the conditions for applying the MVT are not met.

Problem #16

Given the rate function of people entering the museum v(t) = t^2 + \frac{1}{3}t, we want to find the average number of people entering the museum from ( t = 0 ) to ( t = 5 ) hours.

To calculate the total number of people that entered the museum over the 5 hours, we integrate the rate function ( v(t) ) from ( t = 0 ) to ( t = 5 ):

\text{Total number of people} = \int_0^5 v(t) \, dt = \int_0^5 \left( t^2 + \frac{1}{3}t \right) \, dt

\text{Total number of people} = \left[ \frac{t^3}{3} + \frac{t^2}{6} \right]_0^5

\text{Total number of people} = \left( \frac{5^3}{3} + \frac{5^2}{6} \right) - \left( \frac{0^3}{3} + \frac{0^2}{6} \right)

\text{Total number of people} = \frac{125}{3} + \frac{25}{6}

To find the average number of people per hour, we divide the total number by 5 hours:

\text{Average number of people} = \frac{1}{5} \times \text{Total number of people} = \frac{1}{5} \left( \frac{125}{3} + \frac{25}{6} \right)=9.17 \text{ people.}

We can then simplify this expression to get the final average number of people per hour.

Problem #17

To find the average amount of money in an investment account over the first 5 years, given by the function A(t) = 1000e^{0.05t}, we can calculate the integral of ( A(t) ) from ( t=0 ) to ( t=5 ) and then divide by the interval length (5 years).

The setup for the integral is:

\text{Average amount in account} = \frac{1}{5-0} \int_0^5 1000e^{0.05t} \, dt

We make a substitution to simplify the integral:

\text{Let } u = 0.05t \Rightarrow du = 0.05 \, dt \Rightarrow dt = \frac{du}{0.05}

Hence

\int_0^5 1000e^{0.05t} \, dt = 20000 \int_0^{0.25} e^u \, du

The integral becomes:

20000 \int_0^{0.25} e^u \, du = 20000 [e^u]_0^{0.25} = 20000 (e^{0.25} - 1)

Finally, we divide this by 5 to find the average:

\text{Average amount} = \frac{20000 (e^{0.25} - 1)}{5} = 4000 (e^{0.25} - 1)= 1136.10

Problem #18

For the functions f(x) = x^4 and g(x) = x^2 on the interval [1, 2], we will check the validity of Cauchy’s Mean Value Theorem (CMVT). The CMVT states that for two functions ( f(x) ) and ( g(x) ) that are continuous on ([a, b]) and differentiable on ((a, b)), there exists some ( c ) in ((a, b)) such that:

\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{(g(b) - g(a)}

Both ( x^4 ) and ( x^2 ) are continuous and differentiable over the entire real line, satisfying the initial conditions of CMVT.

The derivatives of ( f(x) ) and ( g(x) ) are:

f'(x) = 4x^3

g'(x) = 2x

The values of ( f ) and ( g ) at the endpoints are:

f(1) = 1, \quad f(2) = 16, \quad g(1) = 1, \quad g(2) = 4

Applying the CMVT, we find ( c ) such that:

4c^3 / 2c = (16 - 1)/(4 - 1)

Simplifying:

2c^2 = 15/3

2c^2 = 5

Hence, ( c ) is:

c = \sqrt[2]{\frac{5}{2} }

This value of c falls within the interval ((1, 2)), confirming the CMVT holds for these functions on the given interval.

About Me

My name is Joshua Lizardi. For the past seven years, I have worked for various institutions teaching a wide range of courses in Math, Statistics and Technology. These included Quantitative Reasoning, Calculus ,Applied Technical Mathematics, Remedial Mathematics, Statistics, Computers & Office Automation, Introductory College Algebra, Intermediate College Algebra, Remedial Mathematics, Business Statistics.

I hold a bachelor’s in mathematics (Mercy College), a master’s in applied mathematics (Purdue University), and a master’s in data analytics (Western Governors University).  I also hold a few certifications including  “SAS Certified Statistical Business Analyst SAS 9”, “SAS Certified Base Programmer SAS 9”, “Oracle Database SQL Certified Associate”. 

Subjects like mathematics, statistics, and computer science should not be taught as if they were spectator sports, the best way to learn these subjects is to perform them. Although understanding textbooks and lecture notes is valuable, the learning that comes from one’s own attempts at solving problems is the key to becoming competent in the subject overall. I have always been passionate about mathematics statistics and computer science, and I enjoy encouraging students to see the utility of these subjects. 

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