Question

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

The correct option is A $x=1$ is a local minima and $x=-\frac{2}{3}$ is a local maxima of $f.$

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

$f\left(x\right)=(3x^2-7x+5)e^x$
$f^{\prime }\left(x\right)=(6x-7)e^x+(3x^2-7x+5)e^x$
$=e^x(3x^2-x-2)$
$f^{\prime }\left(x\right)=0$
$\Rightarrow 3x^2-3x+2x-2=0$
$\Rightarrow (3x+2)(x-1)=0$
$\Rightarrow x=1,-\frac{2}{3}$


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