Question

Given that f is even function and g is odd function, how do you determine whether h is even or odd or neither `h(x)=2f(x)+xg(x)`?

Answer 1

Even. See illustrative Socratic graphs, for sample functions.

Explanation:

As f(x) is even, `f(-x) = f(x)`.

As g(x) is odd, `g(-x) = -g(x)`. Now,

`h(-x) = 2f(-x)+(-x)g(-x)=2f(x)+(-x)(-g(x))=2f(x)+xg(x)=h(x)`.

So, `h` is an even function.

Illustrative graph for sample functions+

`f = cos x` and `g = sin x`. Then `h(x)=cosx+xsinx`

See separate graphs for f, g and h.

graph{cosx [-10, 10, -5, 5]}
Symmetrical about y-axis

graph{sinx [-10, 10, -5, 5]}
Symmetrical about origin

graph{2cosx+xsinx [-50, 50 -25, 25]}
Symmetrical about y-axis.