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Chapter 4: Problem 64

Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$g(x)=\sqrt[3]{x-4}$$

Short Answer

Expert verified
Answer: The function \(g(x) = \sqrt[3]{x-4}\) is concave down on the interval \((-\infty, \infty)\), and there are no inflection points.

Step by step solution

01

Find the first derivative of g(x)

To find the first derivative, apply the power rule for derivatives: $$g'(x) = \frac{d}{dx}(\sqrt[3]{x-4}) = \frac{1}{3}(x-4)^{-\frac{2}{3}}.$$
02

Find the second derivative of g(x)

Again, use the power rule for derivatives to find the second derivative: $$g''(x) = \frac{d}{dx}\bigg(\frac{1}{3}(x-4)^{-\frac{2}{3}}\bigg) = -\frac{2}{9}(x-4)^{-\frac{5}{3}}.$$
03

Analyze the sign of the second derivative

To determine the concavity, we need to analyze the sign of the second derivative, \(g''(x)\): $$g''(x) = -\frac{2}{9}(x-4)^{-\frac{5}{3}}.$$ Since \(-\frac{5}{3}\) is a negative exponent, the expression \((x-4)^{-\frac{5}{3}}\) will always be positive, and thus, \(g''(x)\) will always be negative.
04

Determine the intervals of concavity and find inflection points

Since \(g''(x)\) is always negative, the function is concave down on its entire domain. The domain of \(g(x)\) is \((-\infty, \infty)\) since the cube root function is defined for all real numbers. Therefore, the function is concave down on \((-\infty, \infty)\). Since the concavity never changes, there are no inflection points. So the final result is that the function \(g(x) = \sqrt[3]{x-4}\) is concave down on the interval \((-\infty, \infty)\), and there are no inflection points.

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