Question

Find points of local maxima/ minima of
i) $f\left(x\right)=(2^x-1)(2^x-2{)}^2$

Answer 1

$\begin{array}{c}f\left(x\right)=\left(2^x-1\right){\left(2^x-2\right)}^2\\ \Rightarrow f^{\prime }\left(x\right)=\left(2^x-1\right)\times 2\left(2^x-2\right)\times 2{\log }_2+{\left(2^x-2\right)}^2\cdot 2^x{\log }_2=0\\ \Rightarrow f^{\prime }\left(x\right)2^x\left(2^x-1\right){\log }_2\left[\left(2^x-1\right)\times 2+\left(2^x-2\right)\right]=0\\ \Rightarrow f^{\prime }\left(x\right)2^x\left(2^x-2\right){\log }_2\left[2^{x+1}-2+2^x-2\right]=0\\ \Rightarrow f^{\prime }\left(x\right)\left({3.2}^x-4\right)\left(2^x-2\right)2^x{\log }_2=0\\ 2^x=\frac{4}{3}or{2}^x=2,x=1,2^x=0\\ \Rightarrow f^{\prime }\left(x\right)\left({3.2}^{2x}-2^{x+2}\right)\left(2^x-2\right){\log }_2\\ \Rightarrow f\left(x\right)=\left({\mathrm{5.2.2}}^{2x}{\log }_2-{4.2}^x{\log }_2\right)\left(2^x+2\right)+\left({3.2}^{2x}-2^{x+2}\right)2hx{\log }_2\\ \Rightarrow f\left(x\right)=\left(3.4-4.2\right)\times 2{\left({\log }_2\right)}^2<0,\end{array}$

hence x=1 in point of local maxima