If x = 1 is a critical point of the function | vedclass

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(A) Given $f(x) = (3x^{2} + ax - 2 - a)e^{x}$.Differentiating with respect to $x$ using the product rule:$f'(x) = (6x + a)e^{x} + (3x^{2} + ax - 2 - a

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

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(A) Given $f(x) = (3x^{2} + ax - 2 - a)e^{x}$.Differentiating with respect to $x$ using the product rule:$f'(x) = (6x + a)e^{x} + (3x^{2} + ax - 2 - a)e^{x}$$f'(x) = e^{x}(3x^{2} + (6 + a)x - 2)$Since $x = 1$ is a critical point,$f'(1) = 0$:$e^{1}(3(1)^{2} + (6 + a)(1) - 2) = 0$$3 + 6 + a - 2 = 0$$7 + a = 0 \implies a = -7$Substituting $a = -7$ into $f'(x)$:$f'(x) = e^{x}(3x^{2} + (6 - 7)x - 2)$$f'(x) = e^{x}(3x^{2} - x - 2)$$f'(x) = e^{x}(3x + 2)(x - 1)$Setting $f'(x) = 0$,we get critical points $x = 1$ and $x = -\frac{2}{3}$.Using the first derivative test:For $x  0$.For $-\frac{2}{3} For $x > 1$,$f'(x) > 0$.Since $f'(x)$ changes from positive to negative at $x = -\frac{2}{3}$,it is a point of local maxima.Since $f'(x)$ changes from negative to positive at $x = 1$,it is a point of local minima.

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

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