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f(x) = 8(x - 4)^{2/3} + 8. (a) find the interval(s) where f(x) is incre…

Question

f(x) = 8(x - 4)^{2/3} + 8. (a) find the interval(s) where f(x) is increasing. (b) find the interval(s) where f(x) is decreasing. (c) find the x - value(s) of all relative maxima of f(x). if there are none, enter none. if there are multiple relative maxima, separate the values with commas. (d) find the x - value(s) of all relative minima of f(x). if there are none, enter none. if there are multiple relative minima, separate the values with commas.

Explanation:

Response

To solve the problem, we analyze the function \( f(x) = 8(x - 4)^{2/3} + 8 \) by examining its derivative.

Step 1: Find the derivative of \( f(x) \)

We use the chain rule. Let \( u = x - 4 \), so \( f(u) = 8u^{2/3} + 8 \). The derivative of \( f(u) \) with respect to \( u \) is \( f'(u) = \frac{16}{3}u^{-1/3} \), and the derivative of \( u \) with respect to \( x \) is \( u' = 1 \). Thus, the derivative of \( f(x) \) with respect to \( x \) is:
\[
f'(x) = \frac{16}{3}(x - 4)^{-1/3} = \frac{16}{3\sqrt[3]{x - 4}}
\]

Step 2: Analyze the sign of \( f'(x) \)

The denominator \( 3\sqrt[3]{x - 4} \) is zero when \( x = 4 \), so \( f'(x) \) is undefined at \( x = 4 \). For \( x > 4 \), \( \sqrt[3]{x - 4} > 0 \), so \( f'(x) > 0 \). For \( x < 4 \), \( \sqrt[3]{x - 4} < 0 \), so \( f'(x) < 0 \).

Part (a): Interval where \( f(x) \) is increasing

Since \( f'(x) > 0 \) when \( x > 4 \), the function is increasing on the interval \( (4, \infty) \).

Part (b): Interval where \( f(x) \) is decreasing

Since \( f'(x) < 0 \) when \( x < 4 \), the function is decreasing on the interval \( (-\infty, 4) \).

Part (c): Relative maxima

A relative maximum occurs where the function changes from increasing to decreasing. However, \( f(x) \) changes from decreasing to increasing at \( x = 4 \), so there are no relative maxima. Thus, the answer is \( \text{NONE} \).

Part (d): Relative minima

A relative minimum occurs where the function changes from decreasing to increasing. This happens at \( x = 4 \), so the \( x \)-value of the relative minimum is \( 4 \).

Final Answers

(a) The interval where \( f(x) \) is increasing is \( \boxed{(4, \infty)} \).
(b) The interval where \( f(x) \) is decreasing is \( \boxed{(-\infty, 4)} \).
(c) The \( x \)-value(s) of all relative maxima is \( \boxed{\text{NONE}} \).
(d) The \( x \)-value(s) of all relative minima is \( \boxed{4} \).

Answer:

To solve the problem, we analyze the function \( f(x) = 8(x - 4)^{2/3} + 8 \) by examining its derivative.

Step 1: Find the derivative of \( f(x) \)

We use the chain rule. Let \( u = x - 4 \), so \( f(u) = 8u^{2/3} + 8 \). The derivative of \( f(u) \) with respect to \( u \) is \( f'(u) = \frac{16}{3}u^{-1/3} \), and the derivative of \( u \) with respect to \( x \) is \( u' = 1 \). Thus, the derivative of \( f(x) \) with respect to \( x \) is:
\[
f'(x) = \frac{16}{3}(x - 4)^{-1/3} = \frac{16}{3\sqrt[3]{x - 4}}
\]

Step 2: Analyze the sign of \( f'(x) \)

The denominator \( 3\sqrt[3]{x - 4} \) is zero when \( x = 4 \), so \( f'(x) \) is undefined at \( x = 4 \). For \( x > 4 \), \( \sqrt[3]{x - 4} > 0 \), so \( f'(x) > 0 \). For \( x < 4 \), \( \sqrt[3]{x - 4} < 0 \), so \( f'(x) < 0 \).

Part (a): Interval where \( f(x) \) is increasing

Since \( f'(x) > 0 \) when \( x > 4 \), the function is increasing on the interval \( (4, \infty) \).

Part (b): Interval where \( f(x) \) is decreasing

Since \( f'(x) < 0 \) when \( x < 4 \), the function is decreasing on the interval \( (-\infty, 4) \).

Part (c): Relative maxima

A relative maximum occurs where the function changes from increasing to decreasing. However, \( f(x) \) changes from decreasing to increasing at \( x = 4 \), so there are no relative maxima. Thus, the answer is \( \text{NONE} \).

Part (d): Relative minima

A relative minimum occurs where the function changes from decreasing to increasing. This happens at \( x = 4 \), so the \( x \)-value of the relative minimum is \( 4 \).

Final Answers

(a) The interval where \( f(x) \) is increasing is \( \boxed{(4, \infty)} \).
(b) The interval where \( f(x) \) is decreasing is \( \boxed{(-\infty, 4)} \).
(c) The \( x \)-value(s) of all relative maxima is \( \boxed{\text{NONE}} \).
(d) The \( x \)-value(s) of all relative minima is \( \boxed{4} \).