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The function: $\sqrt[4]{4−4}$, $x≤1$ is supposed to have an inverse function of: $(−^4/4)+1$.

But the puzzle is that the function $(−^4/4)+1$ fails the horizontal line test so how on earth could it be the inverse function of $\sqrt[4]{4−4}$.

Is there some sort of explanation for that? Or that the doesn't have an inverse?

              Green = function: ∜(4−4) 
              Red = function:  (−^4/4)+1

Here's the graph: Graph

HeroZhang001
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Markus Maximus
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2 Answers2

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Just take care of the domains and make the codomains equal to the ranges.

$f: (-\infty,1] \to [0,\infty)$ with $f(x)= \sqrt[4]{4-4x}$

has the inverse

$g: [0,\infty) \to (-\infty,1]$ with $g(y)= 1-\frac{x^4}{4}$

Henry
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The function $$ \begin{array}{cccc} f:&(-\infty,1]&\longrightarrow&\mathbb{R}\\ &x&\mapsto&\sqrt[4]{4-4x} \end{array} $$ is injective (since it is strictly increasing) and its range is $[0,+\infty)$. Therefore, it has an inverse $g:[0,+\infty)\longrightarrow(-\infty,1]$. In order to get an analytic expression for $g$, observe that \begin{align} f(x)=y&\iff\sqrt[4]{4-4x}=y\\ &\iff 4-4x=y^4\\ &\iff x=1-\frac{y^4}4. \end{align} So, $\displaystyle g(x)=1-\frac{x^4}4$. The graph of $g$ is the right half of your red graph, which doesn't fail the horizontal line test.

Another User
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  • What I understood is that with inverse functions I only care about the part that reflects my original function especially when changing the y-x's and solving for y gives me a resulting function. That was my mistake. – Markus Maximus Feb 17 '23 at 17:34