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Chapter 1: Problem 62

Show that the sum of two even functions (with the same domain) is an even function.

Short Answer

Expert verified
Given two even functions g(x) and h(x) with the same domain, we create a new function F(x) = g(x) + h(x). To prove this new function is even, we evaluate F(-x) and use the even property of g(x) and h(x). We find that F(-x) = g(x) + h(x), which is equal to F(x). Therefore, the sum of two even functions is also an even function.

Step by step solution

01

Define Even Function

An even function, f(x), is defined by the property that f(x) = f(-x) for all x in its domain. Essentially, this means that the function is symmetric about the y-axis.
02

Let Two Functions be Even

Let's consider two even functions, g(x) and h(x), with the same domain. By definition, they satisfy the following conditions: 1. g(x) = g(-x) 2. h(x) = h(-x)
03

Create a New Function

Now, let's create a new function, F(x), which is the sum of g(x) and h(x): F(x) = g(x) + h(x) Our goal is to prove that F(x) is also an even function.
04

Evaluate F(-x)

To show that F(x) is even, we need to prove that F(x) = F(-x) for all x in its domain. Let's evaluate F(-x): F(-x) = g(-x) + h(-x)
05

Use Even Property of g(x) and h(x)

Since g(x) and h(x) are even functions, we can replace g(-x) with g(x) and h(-x) with h(x): F(-x) = g(x) + h(x)
06

Compare F(x) and F(-x)

Now, we can see that F(-x) is equal to F(x) using our definition: F(x) = g(x) + h(x) F(-x) = g(x) + h(x)
07

Conclusion

Since F(x) = F(-x) for all x in its domain, we have proven that the sum of two even functions is also an even function.

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Most popular questions from this chapter

Explain why an even function whose domain contains a nonzero number cannot be a one-to-one function.

Find functions \(f, g,\) and \(h,\) each simpler than the given function \(T,\) such that \(T=f \circ g \circ h .\) \( \quad T(x)=\sqrt{4+x^{2}}\)

Check your answer by evaluating the appropriate function at your answer. Suppose \(g(x)=\frac{x-3}{x-4}\). Evaluate \(g^{-1}\) (2).

For each of the functions \(f\) given. (a) Find the domain of \(f\). (b) Find the range of \(f\). (c) Find a formula for \(f^{-1}\). (d) Find the domain of \(f^{-1}\). (e) Find the range of \(f^{-1}\). You can check your solutions to part (c) by verifying that \(f^{-1} \circ f=I\) and \(f \circ f^{-1}=I\) (recall that \(I\) is the function defined by \(I(x)=x\) ). \(f(x)=2 x-7\)

For each of the functions \(f\) given. (a) Find the domain of \(f\). (b) Find the range of \(f\). (c) Find a formula for \(f^{-1}\). (d) Find the domain of \(f^{-1}\). (e) Find the range of \(f^{-1}\). You can check your solutions to part (c) by verifying that \(f^{-1} \circ f=I\) and \(f \circ f^{-1}=I\) (recall that \(I\) is the function defined by \(I(x)=x\) ). \(f(x)=\left\\{\begin{array}{ll}2 x & \text { if } x<0 \\ x^{2} & \text { if } x \geq 0\end{array}\right.\) xt { if } x<0 \\ x^{2} & \text { if } x \geq 0\end{array}\right.$
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