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Chapter 2: Problem 112

Suppose \(m\) is an odd integer. Show that the function \(f\) defined by \(f(x)=x^{m}\) is an odd function.

Short Answer

Expert verified
To show that the function \(f(x) = x^m\) is an odd function when \(m\) is an odd integer, we need to prove that \(f(-x) = -f(x)\) for all values of x. By substituting \(-x\) into the function and simplifying, we get \(f(-x) = -1 \cdot x^{2n + 1}\), which is equal to \(-x^m = -f(x)\). Therefore, \(f(x) = x^m\) is an odd function whenever \(m\) is an odd integer.

Step by step solution

01

Write the function with the given definition

We have the function \(f(x) = x^m\), where \(m\) is an odd integer.
02

Find \(f(-x)\)

Now, let's find the value of \(f(-x)\). To do this, we'll substitute \(-x\) into the function: \[ f(-x) = (-x)^m \]
03

Use properties of odd exponents

Since \(m\) is odd, we can write it as \(m = 2n + 1\) for some integer n. Now replace \(m\) in the equation: \[ f(-x) = (-x)^{2n + 1} \]
04

Simplify and compare with \(f(x)\)

We can simplify this equation further: \[ f(-x) = (-1)^{2n + 1} x^{2n+1} \] Now, using the properties of exponents, we see that \((-1)^{2n + 1} = -1\), since the power is odd. So: \[ f(-x) = -1 \cdot x^{2n + 1} \] It turns out that \(x^{2n + 1}\) is equal to \(x^m\), since \(m = 2n + 1\). Therefore, \[ f(-x) = -x^m \] which is equal to \(-f(x)\).
05

Conclusion

Since we have shown that \(f(-x) = -f(x)\) for all x, we can conclude that the function \(f(x) = x^m\) is an odd function whenever \(m\) is an odd integer.

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