The number of points at which the function f( | vedclass

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(A) By the Fundamental Theorem of Calculus,the derivative of the function is:$f'(x) = e^{x-3}(x^2+2)(x-3)(x+4)^2$To find the critical points,set $f'(x

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(A) By the Fundamental Theorem of Calculus,the derivative of the function is:$f'(x) = e^{x-3}(x^2+2)(x-3)(x+4)^2$To find the critical points,set $f'(x) = 0$:$e^{x-3}(x^2+2)(x-3)(x+4)^2 = 0$Since $e^{x-3} > 0$ and $x^2+2 > 0$ for all real $x$,the critical points are $x = 3$ and $x = -4$.We analyze the sign of $f'(x)$ around these points:- For $x > 3$,$f'(x) > 0$.- For $-4 - For $x At $x = 3$,$f'(x)$ changes sign from negative to positive,so $f(x)$ has a local minimum at $x = 3$.At $x = -4$,$f'(x)$ does not change sign (it remains negative on both sides),so it is a point of inflection.Thus,there is only $1$ point of local minimum.

The number of points at which the function $f(x) = \int\limits_0^x {{e^{t - 3}}} \left( {{t^2} + 2} \right)\left( {t - 3} \right){\left( {t + 4} \right)^2}dt$ has a local minimum is:

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