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$$ f:\mathbb R^{+}\to [-5,\infty) $$ $$ f(x)=9x^{2}+6x-5 $$ $$ f^{-1}(y)=({\frac{(\sqrt{y+6})-1}{3}}) $$

Now I have to show that $ f $ is invertible with $f^{-1}$

I'm trying to show that $f^{-1}\circ f(x)=x$ and $f\circ f^{-1}=y$

I have done to some extend: $$ f^{-1}\circ f(x)=f^{-1}(f(x)) $$ $$ = f^{-1}(9x^{2}+6x-5) $$ $$ ={\frac{\sqrt{(9x^{2}+6x-5)+6}-1}{3}} $$ $$ ={\frac{\sqrt{9x^{2}+6x+1}-1}{3}} $$

Now from here it is clear that it it won't be equal to $ x $.

Kindly tell me where I'm getting wrong , is my process wrong or there is any calculation mistake.

Singh
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  • You haven't gone wrong at all, and what's more, you last line does in fact equal $x$. Try factorising $9x^2+6x+1$. –  Mar 07 '14 at 04:46

2 Answers2

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It is exactly equal to $x$, because $9x^2+6x+1=(3x+1)^2$

kmitov
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Who says that won't be equal to x? $9x^2+6x+1=(3x+1)^2$. Plugging that in, we see that the last line is equal to x!

William Chang
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