Question

If $f\left( x \right) = \int\limits_0^x {{e^{{t^2}}}\left( {t - 2} \right)\left( {t - 3} \right)dt}$ for all $x \in \left( {0,\infty } \right)$ then,
A. f has local maximum at x = 2
B. f is decreasing on (2,3)
C. there exists some $c \in \left( {0,\infty } \right)$ such that $f''\left( c \right) = 0$
D. f has a local minimum at x=3

Answer 1

Hint: In order to solve this problem we need to find the differentiation of the function and it is found by simply removing the integral sign and putting x in place of t since the limit is 0 to x. then we will check all the conditions present in the options that the point is maxima if it is changing from positive to negative and vice-versa and if the function as a whole is negative in any interval then the function is decreasing in that interval. Knowing this will solve this problem.

Complete step-by-step answer:
The given function is $f\left( x \right) = \int\limits_0^x {{e^{{t^2}}}\left( {t - 2} \right)\left( {t - 3} \right)dt}$ on differentiation above function after putting the limit we get the differentiation of the above function as,
$ \Rightarrow f'\left( x \right) = {e^{{x^2}}}\left( {x - 2} \right)\left( {x - 3} \right)$
If we put the differentiation above equal to zero then we will get the point of minima and maxima.
So, $f'\left( x \right) = 0$ gives us the value of $x = 2$ and $x = 3$.
If we observe the differentiated function we can clearly see that it is changing the sign from positive to negative at the point 2 that is the function is increasing till 2 and then it is decreasing after 2. So, 2 is the point of local maxima.
We can also see that the function has negative value in the interval when
x$ \in $(2, 3).
So, the function is decreasing in the interval (2, 3).
If we put the value after 3 in the function we can observe that the function is decreasing till 3 then increasing so, 3 is the point of local minima.
If we double differentiate the function that is the differentiation of $f'\left( x \right)$ with respect to x then we get,
$
   \Rightarrow \dfrac{{d\left( {f'\left( x \right)} \right)}}{{dx}} = {e^{{x^2}}}\left( {2x} \right)\left( {x - 2} \right)\left( {x - 3} \right) + {e^{{x^2}}}\left( {x - 2} \right) + {e^{{x^2}}}\left( {x - 3} \right) \\
   \Rightarrow \dfrac{{d\left( {f'\left( x \right)} \right)}}{{dx}} = {e^{{x^2}}}\left( {\left( {2x} \right)\left( {x - 2} \right)\left( {x - 3} \right) + 2x - 5} \right) \\
$
The double differentiation of the function is above, but we have point c$ \in $(2, 3) is zero where f’(c)=0.
This is nothing but Rolle's theorem.
Rolle's theorem, in analysis, is a special case of the mean-value theorem of differential calculus. Rolle's theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.

From the above information we can see that all the options A, B, C, D are correct.

Note: When you get to solve such problems you need to know about maxima and minima and how to find them. Equating the first differentiation to zero will give you these points and all the information has been given above the problem. We also need to know Rolle's theorem, in analysis, a special case of the mean-value theorem of differential calculus. Rolle's theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b. Knowing this will solve your problem and will give you the right answer.