Practice Exam: Finding Extrema on an Interval

Instructions: For each function, determine any relative (local) extrema and absolute (global) extrema on the given interval. If no interval is specified, find any relative extrema that exist.

Problem 1

Find the extrema of \( f(x) = x^3 - 6x^2 + 9x + 1 \) on the interval \([-1, 4]\).

Problem 2

Find the relative extrema of \( g(x) = e^{2x - x^2} \) on the entire real line \((-\infty, \infty)\).

Problem 3

Find the extrema of \( h(x) = x \sin x \) on the interval \([0, 2\pi]\).

Problem 4

Determine the extrema of \( k(x) = \frac{2x}{x^2 + 1} \) on the interval \([-2, 2]\).

Problem 5

Find the extrema of \( p(x) = \cos x + \sin x \) on the interval \([0, \pi]\).

Problem 6

Find any relative extrema of \( q(x) = x^4 - 4x^3 + 6x^2 \) without an interval constraint.

Problem 7

Determine the relative extrema of \( f(x) = e^x \cos x \) on the interval \([0, 2\pi]\).

Problem 8

Find the extrema of \( g(x) = x^2 \sin x \) on the interval \([0, \pi]\).

Problem 9

Find the extrema of \( h(x) = \frac{\sin x}{x} \) on the interval \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right] \).

Problem 10

Determine the relative extrema of \( p(x) = 3x^5 - 5x^3 + x \) on the interval \((-\infty, \infty)\).

Solutions

Solution 1

Function: \( f(x) = x^3 - 6x^2 + 9x + 1 \) on \([-1, 4]\).

  1. Differentiate \( f(x) \): \( f'(x) = 3x^2 - 12x + 9 \).
  2. Set \( f'(x) = 0 \): Solve \( 3x^2 - 12x + 9 = 0 \).
  3. Test endpoints \( x = -1 \) and \( x = 4 \) as well as critical points to find extrema on \([-1, 4]\).

Solution 2

Function: \( g(x) = e^{2x - x^2} \).

  1. Differentiate \( g(x) \) using the chain rule: \( g'(x) = e^{2x - x^2}(2 - 2x) \).
  2. Set \( g'(x) = 0 \) and solve for \( x \) to find critical points.
  3. Determine if critical points are maxima or minima by using the second derivative test.

Solution 3

Function: \( h(x) = x \sin x \) on \([0, 2\pi]\).

  1. Differentiate \( h(x) \) using the product rule: \( h'(x) = \sin x + x \cos x \).
  2. Set \( h'(x) = 0 \) to find critical points within the interval \([0, 2\pi]\).
  3. Test endpoints \( x = 0 \) and \( x = 2\pi \) and critical points for absolute extrema on \([0, 2\pi]\).

Solution 4

Function: \( k(x) = \frac{2x}{x^2 + 1} \) on \([-2, 2]\).

  1. Differentiate \( k(x) \) using the quotient rule: \( k'(x) = \frac{2(x^2 + 1) - 2x(2x)}{(x^2 + 1)^2} \).
  2. Simplify \( k'(x) \) and set it equal to zero to find critical points.
  3. Evaluate \( k(x) \) at endpoints \( x = -2 \) and \( x = 2 \) and critical points to find extrema.

Solution 5

Function: \( p(x) = \cos x + \sin x \) on \([0, \pi]\).

  1. Differentiate \( p(x) \): \( p'(x) = -\sin x + \cos x \).
  2. Set \( p'(x) = 0 \) and solve for \( x \) to find critical points within \([0, \pi]\).
  3. Evaluate \( p(x) \) at endpoints \( x = 0 \) and \( x = \pi \) and critical points to determine extrema.

Solution 6

Function: \( q(x) = x^4 - 4x^3 + 6x^2 \).

  1. Differentiate \( q(x) \): \( q'(x) = 4x^3 - 12x^2 + 12x \).
  2. Set \( q'(x) = 0 \) and solve for \( x \) to find critical points.
  3. Use the second derivative test or analyze sign changes around critical points to determine the nature of extrema.

Solution 7

Function: \( f(x) = e^x \cos x \) on \([0, 2\pi]\).

  1. Differentiate \( f(x) \) using the product rule: \( f'(x) = e^x \cos x - e^x \sin x \).
  2. Simplify and set \( f'(x) = 0 \) to find critical points.
  3. Test endpoints \( x = 0 \) and \( x = 2\pi \) and critical points for extrema on \([0, 2\pi]\).

Solution 8

Function: \( g(x) = x^2 \sin x \) on \([0, \pi]\).

  1. Differentiate \( g(x) \): \( g'(x) = 2x \sin x + x^2 \cos x \).
  2. Set \( g'(x) = 0 \) and solve for critical points within \([0, \pi]\).
  3. Evaluate \( g(x) \) at endpoints \( x = 0 \) and \( x = \pi \) and critical points for extrema.

Solution 9

Function: \( h(x) = \frac{\sin x}{x} \) on \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right] \).

  1. Differentiate \( h(x) \) using the quotient rule: \( h'(x) = \frac{x \cos x - \sin x}{x^2} \).
  2. Set \( h'(x) = 0 \) to find critical points within \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right] \).
  3. Check for local extrema within this interval.

Solution 10

Function: \( p(x) = 3x^5 - 5x^3 + x \).

  1. Differentiate \( p(x) \): \( p'(x) = 15x^4 - 15x^2 + 1 \).
  2. Set \( p'(x) = 0 \) and solve for \( x \) to find critical points.
  3. Use the second derivative test or examine sign changes around each critical point to determine relative extrema on \((-\infty, \infty)\).