Question

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

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

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