Question 4b

SCH_CALC L3CALC

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$$f(x)={\left(\frac{1}{r^r}\right)}^{\frac{1}{r+1}}\cdot x^r$$

where $r>0,r\in \mathbb{R}$

$$f^{\prime }(x)=f^{-1}(x)$$ $$f(f^{-1}(x))=x$$

Start with finding $f^{\prime }(x)$

$$f^{\prime }(x)={\left(\frac{1}{r^r}\right)}^{\frac{1}{r+1}}\cdot r\cdot x^{r-1}$$ $$f^{\prime }(x)=(r^{-r}{)}^{\frac{1}{r+1}}\cdot r\cdot x^{r-1}$$ $$f^{\prime }(x)=r^{\frac{-r}{r+1}}\cdot r^1\cdot x^{r-1}$$ $$f^{\prime }(x)=r^{\frac{-r}{r+1}+1}\cdot x^{r-1}$$ $$f^{\prime }(x)=r^{\frac{-r}{r+1}+\frac{r+1}{r+1}}\cdot x^{r-1}$$ $$f^{\prime }(x)=r^{\frac{1}{r+1}}\cdot x^{r-1}$$

Since $f(f^{-1}x)=x⟺f(f^{\prime }(x))=x$ substitute in $f^{\prime }(x)$ into $f(x)$:

$$f(f^{\prime }(x))={\left(\frac{1}{r^r}\right)}^{\frac{1}{r+1}}\cdot {\left(r^{\frac{1}{r+1}}\cdot x^{r-1}\right)}^r=x$$ $$r^{\frac{-r}{r+1}}\cdot r^{\frac{r}{r+1}}\cdot x^{r(r-1)}=x$$ $$r^{\frac{-r}{r+1}+\frac{r}{r+1}}\cdot x^{r(r-1)}=x$$ $$x^{r(r-1)}=x^1$$ $$r(r-1)=1$$ $$r^2-r-1=0$$ $$r=\frac{1\pm \sqrt{5}}{2}$$

But since $r>0$:

$$\boxed{r=\frac{1+\sqrt{5}}{2}}$$