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Numerade Educator

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Problem 21 Easy Difficulty

Find a formula for the inverse of the function.

$ f(x) = 1 + \sqrt{2 + 3x} $

Answer

$y=f(x)=1+\sqrt{2+3 x} \quad(y \geq 1) \quad \Rightarrow \quad y-1=\sqrt{2+3 x} \Rightarrow(y-1)^{2}=2+3 x \Rightarrow(y-1)^{2}-2=3 x \Rightarrow$ $x=\frac{1}{3}(y-1)^{2}-\frac{2}{3} .$ Interchange $x$ and $y: \quad y=\frac{1}{3}(x-1)^{2}-\frac{2}{3} .$ So $f^{-1}(x)=\frac{1}{3}(x-1)^{2}-\frac{2}{3} .$ Note that the domain of $f^{-1}$ is $x>1$

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Video Transcript

Okay, let's find the inverse. So the first thing we would like to do is just call us why instead of F of X and then we're going to switch X and Y interchange, X and y So we have X equals one plus the square root of two plus three y. Now we want to sell for Why So we'll subtract one from both sides and then we'll square both sides and then we'll subtract two from both sides and then we'll divide both sides by three. Okay, Now we can call this F in verse, so f inverse of X is X minus one quantity squared minus two over three.