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)#?

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Answer 1 · 2017-02-07T07:03:57

Even. See illustrative Socratic graphs, for sample functions.

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.