Questions: Perform a first derivative test on the function f(x)=x^(2/3)(x-4) ;[-4,4]. a. Locate the critical points of the given function. b. Use the First Derivative Test to locate the local maximum and minimum values. c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).

Solution Steps

To perform a first derivative test, we first need to find the derivative of the function \( f(x) = x^{\frac{2}{3}}(x-4) \).

Using the product rule, where if \( u(x) = x^{\frac{2}{3}} \) and \( v(x) = (x-4) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \).

First, find \( u'(x) \) and \( v'(x) \):

• \( u(x) = x^{\frac{2}{3}} \) implies \( u'(x) = \frac{2}{3}x^{-\frac{1}{3}} \).
• \( v(x) = x - 4 \) implies \( v'(x) = 1 \).

Now, apply the product rule: \[ f'(x) = \left(\frac{2}{3}x^{-\frac{1}{3}}\right)(x-4) + x^{\frac{2}{3}}(1) \] \[ f'(x) = \frac{2}{3}x^{-\frac{1}{3}}(x-4) + x^{\frac{2}{3}} \]

Simplify the expression for \( f'(x) \): \[ f'(x) = \frac{2}{3}x^{-\frac{1}{3}}(x-4) + x^{\frac{2}{3}} \] \[ = \frac{2}{3}x^{-\frac{1}{3}}x - \frac{8}{3}x^{-\frac{1}{3}} + x^{\frac{2}{3}} \] \[ = \frac{2}{3}x^{\frac{2}{3}} - \frac{8}{3}x^{-\frac{1}{3}} + x^{\frac{2}{3}} \] \[ = \frac{5}{3}x^{\frac{2}{3}} - \frac{8}{3}x^{-\frac{1}{3}} \]

Critical points occur where \( f'(x) = 0 \) or where \( f'(x) \) is undefined.

Set \( f'(x) = 0 \): \[ \frac{5}{3}x^{\frac{2}{3}} - \frac{8}{3}x^{-\frac{1}{3}} = 0 \] Multiply through by \( 3x^{\frac{1}{3}} \) to clear the fractions: \[ 5x - 8 = 0 \] \[ 5x = 8 \] \[ x = \frac{8}{5} \]

Check where \( f'(x) \) is undefined:

• \( f'(x) \) is undefined at \( x = 0 \) because of the term \( x^{-\frac{1}{3}} \).

Thus, the critical points are \( x = \frac{8}{5} \) and \( x = 0 \).

Examine the sign of \( f'(x) \) around the critical points \( x = \frac{8}{5} \) and \( x = 0 \).

• For \( x < 0 \), choose \( x = -1 \): \[ f'(-1) = \frac{5}{3}(-1)^{\frac{2}{3}} - \frac{8}{3}(-1)^{-\frac{1}{3}} = \frac{5}{3} - \frac{8}{3}(-1) = \frac{5}{3} + \frac{8}{3} = \frac{13}{3} > 0 \]

• For \( 0 < x < \frac{8}{5} \), choose \( x = 1 \): \[ f'(1) = \frac{5}{3}(1)^{\frac{2}{3}} - \frac{8}{3}(1)^{-\frac{1}{3}} = \frac{5}{3} - \frac{8}{3} = -\frac{3}{3} = -1 < 0 \]

• For \( x > \frac{8}{5} \), choose \( x = 2 \): \[ f'(2) = \frac{5}{3}(2)^{\frac{2}{3}} - \frac{8}{3}(2)^{-\frac{1}{3}} \] Approximating: \[ \approx \frac{5}{3}(1.5874) - \frac{8}{3}(0.7937) \approx \frac{5 \times 1.5874}{3} - \frac{8 \times 0.7937}{3} \approx 2.6457 - 2.1163 = 0.5294 > 0 \]

From the sign changes:

• \( x = 0 \) is a local maximum.
• \( x = \frac{8}{5} \) is a local minimum.

Evaluate \( f(x) \) at the critical points and endpoints of the interval \([-4, 4]\).

• \( f(-4) = (-4)^{\frac{2}{3}}(-4-4) = 4 \times (-8) = -32 \)
• \( f(0) = 0^{\frac{2}{3}}(0-4) = 0 \)
• \( f\left(\frac{8}{5}\right) = \left(\frac{8}{5}\right)^{\frac{2}{3}}\left(\frac{8}{5} - 4\right) \) \[ \approx 1.5157 \times \left(\frac{8}{5} - \frac{20}{5}\right) = 1.5157 \times \left(-\frac{12}{5}\right) \approx -3.6377 \]
• \( f(4) = 4^{\frac{2}{3}}(4-4) = 0 \)

The absolute maximum value is \( 0 \) at \( x = 0 \) and \( x = 4 \), and the absolute minimum value is \(-32\) at \( x = -4 \).

Final Answer