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Chapter 11: Problem 62

Find a function \(f\) that satisfies the conditions. $$ f^{\prime \prime}(x)=x^{-3 / 2}, \quad f^{\prime}(1)=2, \quad f(9)=-4 $$

Short Answer

Expert verified
The function that satisfies the provided conditions and its second derivative is \(f(x) = 4x^{1/2} + 2x - 22\).

Step by step solution

01

First Integration

Start by computing the first integral of \(f^{\prime \prime}(x)=x^{-3 / 2}\). The antiderivative of \(x^{-3 / 2}\) is \(-2x^{-1/2}\), so we have \(f^{\prime}(x) = -2x^{-1/2} + C\), where \(C\) is the constant of integration.
02

Calculating the Constant in Step 1

To calculate the constant \(C\), use the given condition \(f^{\prime}(1) = 2\). When we set \(x = 1\) in \(f^{\prime}(x) = -2x^{-1/2} + C\), we get \(C = 2\). Therefore, the first derivative of \(f\) is \(f^{\prime}(x) = -2x^{-1/2} + 2\).
03

Second Integration

Next, compute the second integral using the derivative \(f^{\prime}(x) = -2x^{-1/2} + 2\). The antiderivative of \(-2x^{-1/2}\) is \(4x^{1/2}\) and of \(2\) is \(2x\). Therefore, we get \(f(x) = 4x^{1/2} + 2x + D\), where \(D\) is another constant of integration.
04

Calculating the Constant in Step 3

To calculate the constant \(D\), use the given condition \(f(9) = -4\). Setting \(x = 9\) in \(f(x) = 4x^{1/2} + 2x + D\), we get \(D = -22\).

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Most popular questions from this chapter

Use the Midpoint Rule with \(n=4\) to approximate the area of the region bounded by the graph of \(f\) and the \(x\) -axis over the interval. Compare your result with the exact area. Sketch the region. $$ f(x)=x^{2}-x^{3} $$ $$ [0,1] $$

Find the consumer and producer surpluses. $$ p_{1}(x)=50-0.5 x \quad p_{2}(x)=0.125 x $$

Use a graphing utility to graph the function over the interval. Find the average value of the function over the interval. Then find all \(x\) -values in the interval for which the function is equal to its average value. $$ f(x)=2 e^{x} \quad[-1,1] $$

Use integration to find the area of the triangular region having the given vertices. $$ \begin{aligned} &(0,0),(4,0),(4,4) \\ &(0,0),(4,0),(6,4) \end{aligned} $$

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ f(x)=\sqrt[3]{x}, g(x)=x $$
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