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Chapter 11: Problem 60

Find a function \(f\) that satisfies the conditions. $$ f^{\prime \prime}(x)=x^{2}, \quad f^{\prime}(0)=6, \quad f(0)=3 $$

Short Answer

Expert verified
The function is \(f(x) = \frac{1}{12}x^4 + 6x + 3\)

Step by step solution

01

- Integrate the Second Derivative

The second derivative of \(f\) is given as \(f''(x) = x^2\). To find the first derivative \(f'(x)\), integrate \(f''(x)\). The indefinite integral of \(x^2\) is \(\frac{1}{3}x^3 + C_1\), where \(C_1\) is the constant of integration.
02

- Use Given Condition to Solve for First Constant

The value of \(f'(0)\) is given as \(6\). Substitute \(x = 0\) and \(f'(0) = 6\) into the expression for \(f'(x)\) obtained from the previous step, resulting in \(\frac{1}{3}(0)^3 + C_1 = 6\). Solving this, we find that \(C_1 = 6\). Thus, \(f'(x) = \frac{1}{3}x^3 + 6\).
03

- Integrate the First Derivative

Next, integrate \(f'(x)\) to obtain the original function \(f(x)\). The integral of \(\frac{1}{3}x^3 + 6\) with respect to \(x\) is \(\frac{1}{12}x^4 + 6x + C_2\), where \(C_2\) is another constant of integration.
04

- Use Given Condition to Solve for Second Constant

The value of \(f(0)\) is given as \(3\). Substitute \(x = 0\) and \(f(0) = 3\) into the expression for \(f(x)\) from the previous step. This yields \(\frac{1}{12}(0)^4 + 6(0) + C_2 = 3\). Solving this, we get \(C_2 = 3\). Thus, the original function is \(f(x) = \frac{1}{12}x^4 + 6x + 3\).

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Most popular questions from this chapter

The integrand of the definite integral is a difference of two functions. Sketch the graph of each function and shade the region whose area is represented by the integral. $$ \int_{-1}^{1}\left[\left(1-x^{2}\right)-\left(x^{2}-1\right)\right] d x $$

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ f(x)=e^{0.5 x}, g(x)=-\frac{1}{x}, x=1, x=2 $$

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. $$ f(x)=e^{x / 4} \quad[0,4] $$

You are given the rate of investment \(d l / d t\). Find the capital accumulation over a five-year period by evaluating the definite integral Capital accumulation \(=\int_{0}^{5} \frac{d l}{d t} d t\) where \(t\) is the time in years. $$ \frac{d I}{d t}=\frac{12,000 t}{\left(t^{2}+2\right)^{2}} $$

Use the Trapezoidal Rule with \(n=4\) to approximate the definite integral. $$ \int_{0}^{4} \sqrt{1+x^{2}} d x $$
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