AP Calculus BC 5.5 Using the Candidates Test to Determine Absolute (Global) Extrema Study Notes - New Syllabus
AP Calculus BC 5.5 Using the Candidates Test to Determine Absolute (Global) Extrema Study Notes- New syllabus
AP Calculus BC 5.5 Using the Candidates Test to Determine Absolute (Global) Extrema Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- A function’s derivative can be used to understand some behaviors of the function.
Key Concepts:
- Candidates Test to Determine Absolute (Global) Extrema
Candidates Test to Determine Absolute (Global) Extrema
Using the Candidates Test to Determine Absolute (Global) Extrema
The Candidates Test (also called the Closed Interval Method) is used to find the absolute maximum and absolute minimum of a continuous function \( f(x) \) on a closed interval \([a, b]\).
Steps of the Candidates Test:
- Find the derivative \( f'(x) \).
- Find all critical points in the open interval \( (a, b) \) by solving \( f'(x) = 0 \) or where \( f'(x) \) is undefined.
- Evaluate the function \( f(x) \) at all critical points and at the endpoints \( x = a \) and \( x = b \).
- Compare all the function values. The largest is the absolute maximum; the smallest is the absolute minimum.
Example:
Find the absolute maximum and minimum values of the function \( f(x) = x^3 – 3x^2 + 1 \) on the interval \([0, 3]\).
▶️Answer/Explanation
Find the derivative:
\( f'(x) = 3x^2 – 6x \)
Set \( f'(x) = 0 \):
\( 3x^2 – 6x = 0 \)
\( 3x(x – 2) = 0 \) ⇒ Critical points: \( x = 0 \), \( x = 2 \)
Evaluate \( f(x) \) at all candidates:
\( f(0) = 0^3 – 3(0)^2 + 1 = 1 \)
\( f(2) = 2^3 – 3(2)^2 + 1 = 8 – 12 + 1 = -3 \)
\( f(3) = 3^3 – 3(3)^2 + 1 = 27 – 27 + 1 = 1 \)
Compare values:
Absolute Minimum: \( -3 \) at \( x = 2 \)
Absolute Maximum: \( 1 \) at \( x = 0 \) and \( x = 3 \)
Example :
Find the absolute maximum and minimum of \( f(x) = x\sqrt{4 – x} \) on the interval \([0, 4]\).
▶️Answer/Explanation
The domain of \( f(x) \) is \( [0, 4] \) since \( \sqrt{4 – x} \) is defined only when \( x \leq 4 \).
Find critical points using product rule:
Let \( f(x) = x(4 – x)^{1/2} \) Then, \( f'(x) = \sqrt{4 – x} + \frac{-x}{2\sqrt{4 – x}} \)
Set \( f'(x) = 0 \):
\( \sqrt{4 – x} – \frac{x}{2\sqrt{4 – x}} = 0 \Rightarrow 2(4 – x) = x \Rightarrow 8 – 2x = x \Rightarrow x = \frac{8}{3} \)
Evaluate \( f(x) \) at: – \( x = 0 \): \( f(0) = 0 \) – \( x = \frac{8}{3} \):
\( f\left(\frac{8}{3}\right) \approx \frac{8}{3} \sqrt{4 – \frac{8}{3}} = \frac{8}{3} \sqrt{\frac{4}{3}} = \frac{8}{3} \cdot \frac{2}{\sqrt{3}} = \frac{16}{3\sqrt{3}} \approx 3.08 \) – \( x = 4 \): \( f(4) = 4 \cdot \sqrt{0} = 0 \)
Conclusion: Absolute max ≈ 3.08 at \( x = \frac{8}{3} \), Absolute min = 0 at \( x = 0 \) and \( x = 4 \).
Example :
Find the absolute maximum and minimum of \( f(x) = \frac{x}{x^2 + 1} \) on the interval \([-2, 3]\).
▶️Answer/Explanation
\( f'(x) = \frac{(x^2 + 1)(1) – x(2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 – 2x^2}{(x^2 + 1)^2} = \frac{1 – x^2}{(x^2 + 1)^2} \)
Numerator = 0: \( 1 – x^2 = 0 \Rightarrow x = \pm1 \)
Evaluate at: – \( x = -2 \): \( f(-2) = \frac{-2}{4 + 1} = -\frac{2}{5} \) – \( x = -1 \): \( f(-1) = \frac{-1}{2} \) – \( x = 1 \): \( f(1) = \frac{1}{2} \) – \( x = 3 \): \( f(3) = \frac{3}{10} \)
Conclusion: Absolute max = \( \frac{1}{2} \) at \( x = 1 \) Absolute min = \( -\frac{1}{2} \) at \( x = -1 \)
Example :
The velocity (in m/s) of a particle moving along a line is given by \( v(t) = \sin(t) + \cos(t) \). Find the time in \( [0, 2\pi] \) when velocity is maximum and minimum.
▶️Answer/Explanation
Let \( v(t) = \sin(t) + \cos(t) \) 1. Derivative: \( v'(t) = \cos(t) – \sin(t) \)
\( v'(t) = 0 \): \( \cos(t) = \sin(t) \Rightarrow \tan(t) = 1 \Rightarrow t = \frac{\pi}{4}, \frac{5\pi}{4} \)
Evaluate \( v(t) \) at: – \( v(0) = 0 + 1 = 1 \) – \( v\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.41 \) – \( v\left(\frac{5\pi}{4}\right) = \sin\left(\frac{5\pi}{4}\right) + \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} – \frac{\sqrt{2}}{2} = -\sqrt{2} \approx -1.41 \) – \( v(2\pi) = \sin(2\pi) + \cos(2\pi) = 0 + 1 = 1 \)
Conclusion: Absolute max ≈ \( 1.41 \) at \( t = \frac{\pi}{4} \) Absolute min ≈ \( -1.41 \) at \( t = \frac{5\pi}{4} \)