I have already got $$y=\exp(x^2-x+16)$$ $$\ln y = x^2-x+6$$ $$\ln x=y^2-y+6$$
I know for getting inverse function we need to solve for $x$, but what should i do in this case?
2026-04-04 16:21:34.1775319694
Let $f(x) = \exp (x^2 − x + 6)$. Choose Dom(f) so that $f^{−1}$ exists. What is $f^{−1}$ and Dom($f^{−1}$) in your case?
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(i) We need to solve $$\log y = x^2 - x + 6 \, \Leftrightarrow \, x^2 - x + 6 - \log y = 0.$$ Use the quadratic formula with $a = 1, \, b = -1, \, c = 6 - \log y$.
(ii) For any function $f$, and its inverse $f^{-1}$, we have $\operatorname{dom} f^{-1} = \operatorname{range} f$. This makes sense because the definition of an inverse function is $f^{-1}(f(x)) = x$.