Questions: Let f(x)=4(x-2)^(2 / 3)+8. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima). 1. f is increasing on the intervals 2. f is decreasing on the intervals 3. The relative maxima of f occur at x= 4. The relative minima of f occur at x=

Solution Steps

To solve the problem, we need to analyze the function \( f(x) = 4(x-2)^{2/3} + 8 \) to determine where it is increasing or decreasing and find any relative maxima or minima.

To determine where the function is increasing or decreasing, we first find the derivative \( f'(x) \).

The function is \( f(x) = 4(x-2)^{2/3} + 8 \).

Using the chain rule, the derivative is:

\[ f'(x) = 4 \cdot \frac{2}{3}(x-2)^{-1/3} \cdot 1 = \frac{8}{3}(x-2)^{-1/3} \]

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

1. \( f'(x) = 0 \) implies:

\[ \frac{8}{3}(x-2)^{-1/3} = 0 \]

This equation has no solution because a non-zero constant divided by any real number cannot be zero.

2. \( f'(x) \) is undefined when \( (x-2)^{-1/3} \) is undefined, which occurs at \( x = 2 \).

Thus, the critical point is at \( x = 2 \).

To determine where \( f(x) \) is increasing or decreasing, we test intervals around the critical point \( x = 2 \).

• For \( x < 2 \), choose \( x = 1 \): \[ f'(1) = \frac{8}{3}(1-2)^{-1/3} = \frac{8}{3}(-1)^{-1/3} = -\frac{8}{3} \] Since \( f'(1) < 0 \), \( f(x) \) is decreasing on \( (-\infty, 2) \).

• For \( x > 2 \), choose \( x = 3 \): \[ f'(3) = \frac{8}{3}(3-2)^{-1/3} = \frac{8}{3}(1)^{-1/3} = \frac{8}{3} \] Since \( f'(3) > 0 \), \( f(x) \) is increasing on \( (2, \infty) \).

Since \( f(x) \) changes from decreasing to increasing at \( x = 2 \), there is a relative minimum at \( x = 2 \).

Final Answer