Question

If \[f\left( x \right) = \int_0^x {{e^{{t^2}}}\left( {t - 2} \right)\left( {t - 3} \right)dt} \] for all \[x \in \left( {0,\infty } \right)\]. Which of the following is /are true?
A. \[f\] has a local maximum at \[x = 2\].
B. \[f\] is decreasing on \[\left( {2,3} \right)\].
C. There exists some \[c \in \left( {0,\infty } \right)\] such that \[f''\left( c \right) = 0\]
D. \[f\] has a local minimum at \[x = 3\]
E. All

Answer 1

Hint: First we find the \[f'\left( x \right)\] from the integration. Then we will calculate the second-order derivative of \[f\left( x \right)\]. Then find the critical point by equating \[f'\left( x \right)\] with zero. Then apply the maxima and minima concepts to find the local maximum and local minimum. Then we will apply Rolle’s theorem on \[f'\left( x \right)\].

Formula used:
The derivative of the integration \[g\left( x \right) = \int_0^x {f\left( t \right)dt} \] is \[g'\left( x \right) = f\left( x \right)\].
If \[f''\left( x \right) > 0\] at \[x = a\], then the function \[f\left( x \right)\] has local minima at \[x = a\].
If \[f''\left( x \right) < 0\] at \[x = a\], then the function \[f\left( x \right)\] has local maxima at \[x = a\].
Rolle’s Theorem: If a function \[f\] is continuous on the closed interval \[\left[ {a,b} \right]\]and differentiable on the open interval \[\left( {a,b} \right)\] such that \[f\left( a \right) = f\left( b \right)\], then \[f'\left( x \right) = 0\] for some \[x\] with \[a \le x \le b\].

Complete step by step solution:
Given integration is \[f\left( x \right) = \int_0^x {{e^{{t^2}}}\left( {t - 2} \right)\left( {t - 3} \right)dt} \].
We have to find the derivative of \[f\left( x \right)\].
\[f'\left( x \right) = {e^{{x^2}}}\left( {x - 2} \right)\left( {x - 3} \right)\]
Again, differentiate the above equation with respect to x.
\[f''\left( x \right) = \dfrac{d}{{dx}}\left[ {{e^{{x^2}}}\left( {x - 2} \right)\left( {x - 3} \right)} \right]\]
\[ \Rightarrow f''\left( x \right) = {e^{{x^2}}}\left( {x - 2} \right)\dfrac{d}{{dx}}\left( {x - 3} \right) + {e^{{x^2}}}\left( {x - 3} \right)\dfrac{d}{{dx}}\left( {x - 2} \right) + \left( {x - 2} \right)\left( {x - 3} \right)\dfrac{d}{{dx}}\left[ {{e^{{x^2}}}} \right]\]
\[ \Rightarrow f''\left( x \right) = {e^{{x^2}}}\left( {x - 2} \right) + {e^{{x^2}}}\left( {x - 3} \right) + \left( {x - 2} \right)\left( {x - 3} \right){e^{{x^2}}}\dfrac{d}{{dx}}\left[ {{x^2}} \right]\]
\[ \Rightarrow f''\left( x \right) = {e^{{x^2}}}\left( {x - 2} \right) + {e^{{x^2}}}\left( {x - 3} \right) + \left( {x - 2} \right)\left( {x - 3} \right){e^{{x^2}}}\left( {2x} \right)\]
Simplify the above equation:
\[ \Rightarrow f''\left( x \right) = {e^{{x^2}}}\left[ {\left( {x - 2} \right) + \left( {x - 3} \right) + 2x\left( {x - 2} \right)\left( {x - 3} \right)} \right]\]
\[ \Rightarrow f''\left( x \right) = {e^{{x^2}}}\left[ {x - 2 + x - 3 + 2x\left( {{x^2} - 5x + 6} \right)} \right]\]
\[ \Rightarrow f''\left( x \right) = {e^{{x^2}}}\left[ {2x - 5 + 2{x^3} - 10{x^2} + 12x} \right]\]
\[ \Rightarrow f''\left( x \right) = {e^{{x^2}}}\left[ {2{x^3} - 10{x^2} + 14x - 5} \right]\]
Equate \[f'\left( x \right)\] equal to zero to find the critical point.
\[{e^{{x^2}}}\left( {x - 2} \right)\left( {x - 3} \right) = 0\]
Equate each factor with zero
\[\left( {x - 2} \right) = 0\]
\[ \Rightarrow x = 2\]
\[\left( {x - 3} \right) = 0\]
\[ \Rightarrow x = 3\]
The critical points of \[f\left( x \right)\] are 2, 3.
Substitute \[x = 2\] in \[f''\left( x \right) = {e^{{x^2}}}\left[ {2{x^3} - 10{x^2} + 14x - 5} \right]\]
\[ \Rightarrow f''\left( x \right) = {e^{{2^2}}}\left[ {2 \cdot {2^3} - 10 \cdot {2^2} + 14 \cdot 2 - 5} \right]\]
\[ \Rightarrow f''\left( x \right) = - {e^{{2^2}}} < 0\]
Thus \[f\left( x \right)\] has local maximum at \[x = 2\].
Substitute \[x = 3\] in \[f''\left( x \right) = {e^{{x^2}}}\left[ {2{x^3} - 10{x^2} + 14x - 5} \right]\]
\[ \Rightarrow f''\left( x \right) = {e^{{3^2}}}\left[ {2 \cdot {3^3} - 10 \cdot {3^2} + 14 \cdot 3 - 5} \right]\]
\[ \Rightarrow f''\left( x \right) = 11{e^{{2^2}}} > 0\]
Thus \[f\left( x \right)\] has local minima at \[x = 3\].
Check whether the function is increasing or decreasing in \[\left( {2,3} \right)\].
The value of \[\left( {x - 3} \right)\] must be negative when \[x \in \left( {2,3} \right)\].
The value of \[\left( {x - 2} \right)\] must be positive when \[x \in \left( {2,3} \right)\].
The value of \[{e^{{x^2}}}\] must be positive when \[x \in \left( {2,3} \right)\].
So, the product \[{e^{{x^2}}}\left( {x - 2} \right)\left( {x - 3} \right)\] is a negative number.
Thus \[f'\left( x \right) < 0\] for all \[x \in \left( {2,3} \right)\].
So, the function \[f\left( x \right)\] is decreasing in \[x \in \left( {2,3} \right)\].

At \[x = 2\], \[f'\left( x \right) = 0\] and at \[x = 3\], \[f'\left( x \right) = 0\].
So, \[f'\left( 2 \right) = f'\left( 3 \right)\]
Now apply Rolle’s theorem on \[f'\left( x \right)\].
Since \[f'\left( 2 \right) = f'\left( 3 \right)\], there exist \[c \in \left( {2,3} \right)\] such that \[f''\left( c \right) = 0\].
This implies there exist \[c \in \left( {0,\infty } \right)\] such that \[f''\left( c \right) = 0\].
Hence option E is the correct option.

Note: Students often do mark correct option C as they get there exist \[c \in \left( {2,3} \right)\] such that \[f''\left( c \right) = 0\]. But if \[c \in \left( {2,3} \right)\] which means \[c\] also belongs to \[\left( {0,\infty } \right)\]. So option C also correct.