If $\displaystyle h(x)=x^3 +2x+1$ has an inverse $\displaystyle h^{-1}$ on $\displaystyle \mathbb{R}$ , how would you find the value of $\displaystyle (h^{-1})')(y)$ at the points corresponding to $\displaystyle x=0,1,-1$ ?

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- Mar 22nd 2010, 03:44 AMfrenchguy87Polynomials and inverses
If $\displaystyle h(x)=x^3 +2x+1$ has an inverse $\displaystyle h^{-1}$ on $\displaystyle \mathbb{R}$ , how would you find the value of $\displaystyle (h^{-1})')(y)$ at the points corresponding to $\displaystyle x=0,1,-1$ ?

- Mar 22nd 2010, 04:14 AMtonio

By a well-known theorem, $\displaystyle (h^{-1})'=\frac{1}{h'}$ , with the proper variables in each, so in this case $\displaystyle (h^{-1})'(1)=\frac{1}{h'(0)}=\frac{1}{2\cdot 0^2+2}=\frac{1}{2}$ , since $\displaystyle h(0)=1\Longleftrightarrow h^{-1}(1)=0$ ...now you ty to do the other ones.

Tonio