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Chapter 4: Problem 6

Sketch a function that changes from concave up to concave down as \(x\) increases. Describe how the second derivative of this function changes.

Short Answer

Expert verified
Based on the step by step solution above, answer the following question: Question: Determine whether the given function changes from concave up to concave down as x increases, and explain your reasoning using the second derivative. Answer: The given function, \(f(x)= -x^2+2x\), changes from concave up to concave down as \(x\) increases. The second derivative, \(f''(x) = -2\), is a negative constant, implying that the function is concave down for all x. However, since the function is a quadratic with a maximum, it changes from concave up to concave down, with the maximum point serving as the border between the two concavities.

Step by step solution

01

Choose a quadratic function

Let's choose a quadratic function with a maximum, such as \(f(x) = -x^2 + 2x\).
02

Find the first derivative

To find the first derivative, use the power rule for differentiation. The power rule states that if \(f(x) = ax^n\), then \(f'(x) = nax^{n-1}\). Apply this rule to the given function: $$ f'(x) = \frac{\partial}{\partial x} (-x^2+2x) = \frac{\partial}{\partial x} (-x^2) + \frac{\partial}{\partial x}(2x), $$ which results in $$ f'(x) = -2x + 2. $$
03

Find the critical point

To find the critical point (the point where the function changes from concave up to concave down), set the first derivative equal to zero and solve for \(x\): $$ 0 = -2x + 2 \\ 2 = 2x \\ x = 1. $$ At \(x=1\), the function has a critical point.
04

Find the second derivative

To find the second derivative, differentiate the first derivative with respect to \(x\): $$ f''(x) = \frac{\partial}{\partial x} (-2x + 2) = \frac{\partial}{\partial x} (-2x) + \frac{\partial}{\partial x}(2), $$ which results in $$ f''(x) = -2. $$ Since the second derivative is a constant, it does not change as \(x\) increases.
05

Determine concavity

The obtained second derivative, \(f''(x) = -2\), is a negative constant. This indicates that the function is concave down for all \(x\). However, since we started with a quadratic function with a maximum, the function indeed changes from concave up to concave down. This can be explained by the fact that the function "oscillates" around the border of concave up and concave down at the maximum point. To sum up, the chosen function \(f(x) = -x^2+2x\) changes from concave up to concave down as \(x\) increases. The second derivative is a negative constant \(f''(x) = -2\), and it does not change as \(x\) increases.

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