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Numerade Educator

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Problem 42 Medium Difficulty

Find $ f \circ g \circ h $.

$ f(x) = \tan x $ , $ g(x) = \dfrac{x}{x - 1} $ , $ h(x) = \sqrt[3]{x} $

Answer

$f \circ g \circ h=\tan \left[\frac{\sqrt[3]{x}}{\sqrt[3]{x}-1}\right]$

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Video Transcript

here we have the functions F, g and H, and we're going to find the composition f of G of h. Another way to write that is with parentheses, f of G of h of X. And when we read it that way we can see that H is the inside function. It's going to be substituted into G. So we're going to take the cube root of X and we're going to substituted into G. In both places, we see an X that gives us G of h of X. So G of h of X is the cube root of X over the cube root of X minus one. So that whole thing goes inside of death. So we take that whole thing and we substituted in to the F function in place of X. And that's going to result in the tangent of that Hubert of X, over the cube root of X minus one