Given function

$f(x)=(3x^2+ax-2-a)e^x$

We are told that $x=1$ is a critical point, so:

$f'(1)=0$

Differentiate $f(x)$

Let $g(x)=3x^2+ax-2-a$ so that $f(x)=g(x)e^x$

Using product rule, $f'(x)=e^x(g'(x)+g(x))$

Now, $g'(x)=6x+a$

Hence, f'(x)=e^xig[(6x+a)+(3x^2+ax-2-a)ig]

Simplifying, $f'(x)=e^x(3x^2+ax+6x-2)$ $f'(x)=e^x(3x^2+(a+6)x-2)$

Use the critical point condition at $x=1$

Since $f'(1)=0$ and $e^1\neq 0$, we must have $3(1)^2+(a+6)(1)-2=0$ $3+a+6-2=0$ $a+7=0$ $a=-7$

Substitute $a=-7$ into $f'(x)$

$f'(x)=e^x(3x^2+(-7+6)x-2)$ $f'(x)=e^x(3x^2-x-2)$

Factor the quadratic: $3x^2-x-2=(3x+2)(x-1)$

Therefore, $f'(x)=e^x(3x+2)(x-1)$

Since $e^x>0$ for all $x$, the sign of $f'(x)$ depends on $(3x+2)(x-1)$

Find the critical points

$f'(x)=0 \iff (3x+2)(x-1)=0$

So critical points are: $x=1,\quad x=-\frac23$

Sign analysis of $f'(x)$

Check intervals around $x=-\frac23$ and $x=1$:

• For $x<-\frac23$, take $x=-1$: $(3x+2)(x-1)=(-1)(-2)>0$ so $f'(x)>0$

• For $-\frac23<x<1$, take $x=0$: $(2)(-1)<0$ so $f'(x)<0$

• For $x>1$, take $x=2$: $(8)(1)>0$ so $f'(x)>0$

Thus,

• at $x=-\frac23$, $f'(x)$ changes from $+$ to $-$, so $x=-\frac23$ is a local maximum.
• at $x=1$, $f'(x)$ changes from $-$ to $+$, so $x=1$ is a local minimum.

Match with options

This corresponds to:

D: $x=1$ is a local minima and $x=-\frac23$ is a local maxima of $f$.