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How do you find g[h(x)] and h[g(x)] given #h(x)=2x-1# #g(x)=3x+4#?

Answer 1

To find g[h(x)], substitute h(x) into g(x) wherever there is an x. So g[h(x)] = g(2x - 1) = 3(2x - 1) + 4 = 6x - 3 + 4 = 6x + 1.

To find h[g(x)], substitute g(x) into h(x) wherever there is an x. So h[g(x)] = h(3x + 4) = 2(3x + 4) - 1 = 6x + 8 - 1 = 6x + 7.

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Answer 2

Caleb Alexander

g(h(x))=6x +1 and h(g(x)) = 6x+7

Basically , this is a substitution exercise.

g(h(x)) = g(2x-1) . Now substitute x = 2x-1 in for x in g(x)

hence g(2x-1) = 3(2x-1)+4 = 6x - 3 + 4 = 6x + 1

Similarly : h(g(x)) = h(3x+4) substitute x = 3x+4 in h(x)

hence h(3x+4) = 2(3x + 4 ) - 1 = 6x + 8 - 1 = 6x + 7

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.