Question

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

Short answer

The function h is always an even function.

Step-by-step solution

Check the composition for properties of even function

Find the value of \(h\left( { - x} \right)\):

\(\begin{aligned}h\left( { - x} \right) &= f \circ g\left( { - x} \right)\\ &= f\left( {g\left( { - x} \right)} \right)\\ &= f\left( {g\left( x \right)} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {g\,{\rm{is}}\,\,{\rm{even}}} \right)\end{aligned}\)

Check for \(h\left( x \right)\)

As \(h\left( x \right) = h\left( { - x} \right)\), so h is always an even function.