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Chapter 1: Problem 36

Evaluate each function at the given values of the independent variable and simplify. \(f(x)=\frac{4 x^{3}+1}{x^{3}}\) a. \(f(2)\) b. \(f(-2)\) c. \(f(-x)\)

Short Answer

Expert verified
\(f(2) = \frac{33}{8}, f(-2) = \frac{31}{8}, f(-x) = \frac{4x^{3}-1}{x^{3}}\).

Step by step solution

01

Evaluate f(2)

Plug in 2 for x in the function, i.e., \(f(2)=\frac{4*(2)^{3}+1}{(2)^{3}}\). By simplifying this expression, we get \(f(2) = \frac{4*8+1}{8} = \frac{33}{8}\).
02

Evaluate f(-2)

Next, plug in -2 for x in the function, i.e., \(f(-2)=\frac{4*(-2)^{3}+1}{(-2)^{3}}\). By simplifying this expression, we get \(f(-2) = \frac{4*(-8)+1}{-8} = \frac{-32+1}{-8} = \frac{-31}{-8} = \frac{31}{8}\).
03

Evaluate f(-x)

Lastly, substitute -x into the function for x, that is, \(f(-x) = \frac{4*(-x)^{3}+1}{(-x)^{3}}\). Simplifying this expression gives us \(f(-x) = \frac{-4x^{3}+1}{-x^{3}}\). We can simplify further by multiplying the top and bottom by -1. This gives us \(f(-x) = \frac{4x^{3}-1}{x^{3}}\).

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Most popular questions from this chapter

Find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=\frac{1}{2 x}$$

I noticed that the difference quotient is always zero if \(f(x)=c,\) where \(c\) is any constant.

If you are given a function's equation, how do you determine if the function is even, odd, or neither?

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Explaining the Concepts: If equations for two functions are given, explain how to obtain the quotient function and its domain.
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