Question

If $f\left(x\right)=\left(e^{(x-1)}{)}^2\right(x-1{)}^2$, then which of the following is $CORRECT$?

• A. $f\left(x\right)$ has an extremum at $x=\pm 1$
• B. $f^{\prime }\left(1\right)=0$
• C. $f^{\prime \prime }\left(1\right)<0$
• D. $f^{\prime \prime }\left(1\right)>0$

Answer 1

The correct option is B $f^{\prime }\left(1\right)=0$

$\begin{array}{c}f\left(x\right)=(e^{(x-1)}{)}^2(x-1{)}^2\\ \Rightarrow f^{\prime }\left(x\right)=\frac{d\left[e^{2\left(x-1\right)}{(x-1)}^2\right]}{dx}\\ \Rightarrow f^{\prime }\left(x\right)=e^{2\left(x-1\right)}\frac{d{\left(x-1\right)}^2}{dx}+{\left(x-1\right)}^2\frac{d{e}^{2\left(x-1\right)}}{dx}\\ \Rightarrow f^{\prime }\left(x\right)=e^{\left(2x-2\right)}2\left(x-1\right)+{\left(x-1\right)}^2{e}^{2\left(x-1\right)}\\ \Rightarrow f^{\prime }\left(x\right)=2e^{2\left(x-1\right)}\left[\left(x-1\right)+{\left(x-1\right)}^2\right]\\ purx=1\\ f^{\prime }\left(1\right)=2e^{2\left(0\right)}\times \left(0+0\right)=0\end{array}$