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Let $f(x) = (x+2)^2-5$. If the domain of $f$ is all real numbers, then $f$ does not have an inverse function, but if we restrict the domain of $f$ to an interval $[c,\infty)$, then $f$ may have an inverse function. What is the smallest value of $c$ we can use here, so that $f$ does have an inverse function?

Aug 17, 2019

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Let$$f(x) = (x+2)^2-5$$.
If the domain of $$f$$ is all real numbers,
then $$f$$ does not have an inverse function,
but if we restrict the domain of $$f$$ to an interval $$[c,\infty)$$,
then $$f$$ may have an inverse function.
What is the smallest value of $$c$$ we can use here,
so that $$f$$ does have an inverse function?

$$\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x+2)^2-5} \\ \\ x &=& \Big(f^{-1}(x)+2\Big)^2-5 \quad | \quad \text{inverse function} \\ x+5 &=& \Big(f^{-1}(x)+2\Big)^2 \\ \Big(f^{-1}(x)+2\Big)^2 &=& x+5 \\ f^{-1}(x)+2 &=& \pm\sqrt{x+5} \\ \mathbf{f^{-1}(x)} &=& \mathbf{-2 \pm\sqrt{x+5}} \\ \hline \\ \sqrt{x+5} &=& 0 \quad | \quad \text{the smallest value of }x \text{ is } c \\ x+5 &=& 0 \\ x &=& -5 \quad | \quad \text{the smallest value is } -5 \\ \\ \hline \mathbf{[c,\infty)} &=& \mathbf{[-5,\infty)} \\ \hline \end{array}$$ Aug 18, 2019
edited by heureka  Aug 19, 2019