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Numerade Educator

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Problem 16 Easy Difficulty

If $ f(x) = x^5 + x^3 + x $ , find $ f^{-1} (3) $ and $ f (f^{-1} (2)) $.

Answer

$f^{-1}(3)=1$
$f\left[f^{-1}(2)\right]=2$

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Video Transcript

here we have f of X, and we're looking for F inverse of three. Since this is an in vertebral function, we know it's 1 to 1. And so let's suppose that F inverse of three was X. That means that f of X is three because the output of the inverse will be the input of the function and vice versa. So what we're saying here is that we're looking for the X coordinate that has a wide coordinated three so we can go ahead and substitute three into our equation for why Now we know there's only one value of X that makes this true, since it's a 1 to 1 function and we can solve this one by inspection if we just stop and think about it for a minute. We realized that if we had one plus one plus one, we would have three and want to. The fifth is one and want to. The third is one and one is one, and so we can let X equal one for the second part of the problem. We're finding f of f inverse of to. So what that means is the output of F is to what was the input? Now what's the output? So we're finding the same output that we started with, so the answer is to