E1.

Evaluate the function \(f(x) = |x^2 - 8\sqrt{x} -5|\) for a. \(f(9)\), b.\(f(4)\)

Solution

1. \(f(9) = |9^2 - 8\sqrt{9} -5| = |52| = 52\)
2. \(f(4) = |4^2 - 8\sqrt{4} -5| = |-5| = 5\)

E2.

We can plot inequalities on the real line as in these examples

1. Plot on the real line the inequality \(|x - 2| < 3\)

2. What are the boundary points of the interval for \(x\)?

3. What is the meaning of the numbers \(2\) and \(3\) (when looking at the plot)?

In case you need a little hint.

Solution

\((x - 2) < 3\) and \(-(x - 2) < 3\) which means the interval is \(\mathbf{-1} < x < \mathbf{5}\).

\(2\) is the center of the interval and \(3\) is the distance from it on each side (the total length is \(6\)).

E3.

For the following pairs of functions solve for \(x\) the equation \(f(x)=g(x)\) (that is find the solution \(x=?\))

1. \(f(x) = |5-2x|-1, g(x) = 2\)
2. \(f(x) = |2x-6|-3, g(x) = \frac{x}{2}+2\)
3. \(f(x) = 4|x-2|+3, g(x) = -3x + 1\)

In case you need a little hint.

Then use the following graphing tool Geogebra to plot the pairs of functions \(f\) and \(g\) into the same graph (the tool is fairly intuitive, I’m sure you will figure out how to use it). A few hints for Geogebra are here.

What is the relation between the plots and the solutions for \(x\)?

Solution

1. \(f(x) = |5-2x|-1, g(x) = 2\) \[\begin{gather} |5-2x|-1 = 2 \\ \pm (5 - 2x) = 3\\ 5 - 2x = 3 \text{ for } x \leq 5/2 \qquad -5 + 2x = 3 \text{ for } x \geq 5/2 \\ 2x = 2 \qquad 2x = 8 \\ x = 1 \qquad x = 4 \end{gather}\]

2. \(f(x) = |2x-6|-3, g(x) = \frac{x}{2}+2\) \[\begin{gather} |2x-6|-3 = \frac{x}{2}+2 \\ \pm (2x-6) = \frac{x}{2}+5 \\ 2x - \frac{x}{2} = 11 \text{ for } x \geq 3 \qquad -2x - \frac{x}{2} = -1 \text{ for } x \leq 3 \\ 3x = 22 \qquad -5x = -2 \\ x = 22/3 \qquad x = 2/5 \end{gather}\]

3. \(f(x) = 4|x-2|+3, g(x) = -3x + 1\) \[\begin{gather} 4|x-2|+3 = -3x + 1 \\ \pm 4(x-2) = -3x -2 \\ 4x - 8 = -3x - 2 \text{ for } x \geq 2 \qquad -4x +8 = -3x - 2 \text{ for } x \leq 2 \\ 7x = 10 \qquad -x = -10 \\ x = 10/7 \not\geq 2 \text{ not a solution } \qquad x = 10 \not\leq 2 \text{ not a solution } \end{gather}\]

E4.

Determine for which values of \(x\) the following inequalities hold

1. \(-\frac{1}{2}|4x-5| + 3 < 0\)
2. \(-2|x-4| \leq -6\)
3. \(|3x-5| \leq 11\)

Use the graphing tool Geogebra to plot the functions and relate the solutions of your calculations to the plot.

Solution

1. \(-\frac{1}{2}|4x-5| + 3 < 0\) \[\begin{gather} -\frac{1}{2}|4x-5| < -3 \\ |4x-5| > 6 \\ 4x - 5 > 6 \text{ for } x \geq 5/4 \qquad -4x + 5 > 6 \text{ for } x \leq 5/4 \\ x > 11/4 \qquad x < 1/4\\ \end{gather}\]

2. \(-2|x-4| \leq -6\) \[\begin{gather} |x-4| \geq 3 \\ x - 4 \geq 3 \text{ for } x \geq 4 \qquad -x + 4 \geq 3 \text{ for } x \leq 4 \\ x \geq 7 \qquad x \leq 1\\ \end{gather}\]

3. \(|3x-5| \leq 11\) \[\begin{gather} 3x - 5 \leq 11 \text{ for } x \geq 5/3 \qquad -3x + 5 \leq 11 \text{ for } x \leq 5/3 \\ 3x \leq 16 \qquad -3x \leq 6\\ x \leq 16/3 \qquad x \geq -2\\ \end{gather}\]

E5.

Convert these radian measures of an angle \(\theta\) into degrees (e.g. \(\pi/2 = 90^{\circ}\))

1. \(\frac{\pi}{6}\)
2. \(3\)
3. \(\frac{5\pi}{2}\)

In case you need a little hint.

Solution

1. \(\frac{\pi}{6} = \frac{\pi}{6}\frac{180}{\pi} = 30^o\)
2. \(3 = 3\frac{180}{\pi} = 171^o\)
3. \(\frac{5\pi}{2}= \frac{5\pi}{2}\frac{180}{\pi} = 450^o = 360^o + 90^o\)

E6.

Convert these degree measures of an angle \(\theta\) into radians (e.g. \(90^{\circ} = \pi/2\))

1. \(45^{\circ}\)
2. \(-60^{\circ}\)
3. \(210^{\circ}\)

In case you need a little hint.

Solution

1. \(45^{\circ} = 45^{\circ} \frac{\pi}{180} = \frac{\pi}{4}\)
2. \(-60^{\circ} = -60^{\circ} \frac{\pi}{180} = -\frac{\pi}{3} = \frac{5\pi}{3}\)
3. \(210^{\circ} = 210^{\circ} \frac{\pi}{180} = \frac{7\pi}{6}\)

E7.

Use the unit circle (don’t use a calculator) to get the values of the functions \(g(x) = \sin(x)\) and \(h(x) = \cos(x)\) for these values of the angle \(x\)

1. \(\frac{3}{2}\pi\)
2. \(\frac{7\pi}{6}\)
3. \(-\pi\)
4. \(\frac{11\pi}{4}\)

Hints for some special angles: \(\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\), \(\sin(\frac{\pi}{6}) = \frac{1}{2}\), \(\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\). Use the symmetries along the circle to derive the other values.

Solution

1. \(\sin \frac{3}{2}\pi = -1, \quad \cos \frac{3}{2}\pi = 0\)
2. \(\sin \frac{7\pi}{6} = -\frac{1}{2}, \quad \cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2}\)
3. \(\sin -\pi = 0, \quad \cos -\pi = -1\)
4. \(\sin \frac{11\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos \frac{11\pi}{4} = -\frac{\sqrt{2}}{2}\)

E8.

There is an important relationship between the functions \(\sin(\theta)\) and \(\cos(\theta)\) that you should be able to derive yourself when you think carefully about the unit circle and the way we get the values of the two functions. For any angle \(\theta\) we have

\[(\sin \theta)^2 + (\cos \theta)^2 = \ ? \] Replace the question mark in the above with your answer and explain how you derived it.

In case you need a little hint.

Solution

Observe that the \(\sin\) and \(\cos\) form a right angled triangle with hypotenuse of lenght \(1\) and recall the Pythagoras theorem \(c^2 = a^2 + b^2\)

\[(\sin \theta)^2 + (\cos \theta)^2 = \ 1 \]

E9.

Use the graphing tool Geogebra to plot the functions below and find their values for the angles \(x = \pi\) and \(x = 5\)

1. \(f(x) = \cos(x + 5)\)
2. \(f(x) = 3\sin(x)\)
3. \(f(x) = \cos(2x + 4)\)
4. \(f(x) = 4\sin(0.5x -1)\)

Hint for Geogebra: you can put as input \(x = 5\) to draw a vertical line at \(5\).

Solution

E10.

Find the maxima and minima for the functions in E9.

1. What are the maximum \(\max f(x)\) and minimum \(\min f(x)\) values of the functions?
2. At which points \(x\) do the functions have the maxima \(\text{argmax } f(x)\) and minima \(\text{argmin } f(x)\)?

Note the difference between \(\max f(x)\) and \(\text{argmax } f(x)\), we will see this a lot in the course!

Solution

Graphs

There are multiple local minima and maxima because the functions are periodic.

For example for the function \(f(x) = \cos(x + 5)\) one of the maxima is \[\max \cos(x+5) = 1, \qquad \text{argmax} \cos(x+5) = 1.283\] and one of the minima \[\min \cos(x+5) = -1, \qquad \text{argmin} \cos(x+5) = 4.424\] They repeat every \(2\pi\).