Question

Suppose g is an even function and let\(h = f \circ g\). Is h always an even function?

Short answer

Yes, because \(h\left( x \right) = h\left( { - x} \right)\),\(h\) is an even function.

Step-by-step solution

Introduction

Two functions can be expressed as a combination of each other with the addition, subtraction, multiplication, and divide operation. One or more operations can be applied at a time.

If f and g are two functions they can be written as,

\(\left( f+g \right)\left( x \right),\left( f-g \right)\left( x \right),\left( f\circ g \right)\left( x \right),\left( {f}/{g}\; \right)\left( x \right)\)

Given

\(h = f \circ g\)

Function h

We have to find out \(h\left( { - x} \right)\),

\(\begin{array}{l}h( - x) = \left( {f \circ g} \right)\left( { - x} \right)\\h( - x) = f\left( {g\left( { - x} \right)} \right)\\h( - x) = f\left( {g\left( x \right)} \right)\\h( - x) = h(x)\end{array}\)

Here gis even as given.

And because \(h\left( x \right) = h\left( { - x} \right)\),\(h\)is an even function.