Question

If $x=1$ is a critical point of the function $f(x)=\left(3 x^{2}+a x-2-a\right) e^{x},$ then :

• A. $x=1$ is a local minima and $x=-\frac{2}{3}$ is a local maxima of $f$
• B. $x=1$ is a local maxima and $x=-\frac{2}{3}$ is a local minima of $f$
• C. $x=1$ and $x=-\frac{2}{3}$ are local minima of $f$
• D. $x =1$ and $x =-\frac{2}{3}$ are local maxima of $f$

Answer 1

Answer: (A)

$f(x)=\left(3 x^{2}+a x-2-a\right) e^{x}$
$f'(x)=\left(3 x^{2}+a x-2-a\right) e^{x}+e^{x}(6 x+a)$
$=e^{x}\left(3 x^{2}+x(6+a)-2\right)$
$f'(x)=0$ at $x=1$
$\Rightarrow 3+(6+a)-2=0$
$a=-7$
$f'(x)=e^{x}\left(3 x^{2}-x-2\right)$
$=e^{x}(x-1)(3 x+2)$

$x =1$ is point of local minima
$x =\frac{-2}{3}$ is point of local maxima