Absolute maximum value of the function (x^2 – 2x + 7)e^(4x^3 – 12x^2 – 180x + 31)

Maths Tutor

2 years ago

If the absolute maximum value of the function

$f(x) = \left( {{x^2} - 2x + 7} \right){e^{\left( {4{x^3} - 12{x^2} - 180x + 31} \right)}} \text{ in the interval [-3, 0] }is f(\alpha )$ , then

$\alpha = 0$
$\alpha = - 3$
$\alpha \in ( - 1,0)$
$\alpha \in ( - 3, - 1]$

Solution:

Tip for solving this question:

Find differentiation of f(x)

The absolute maximum value of x

Step 1 of 2:

Given, $f(x) = \left( {{x^2} - 2x + 7} \right){e^{\left( {4{x^3} - 12{x^2} - 180x + 31} \right)}}$

Take $f(x) = \underbrace {\left( {{x^2} - 2x + 7} \right)}_{{f_1}(x)}\underbrace {{e^{\left( {4{x^3} - 12{x^2} - 180x + 31} \right)}}}_{{f_2}(x)}$

Now, differentiate ${f_1}(x)$

$\begin{gathered} \Rightarrow f_1'(x) = \frac{d}{{dx}}\left( {{x^2} - 2x + 7} \right) \hfill \\ \Rightarrow f_1'(x) = 2x - 2 \hfill \\ \end{gathered}$

For critical points,

${f'}(x) = 0 \hfill \\ \Rightarrow 2x - 2 = 0 \hfill \\ \Rightarrow 2x = 2 \hfill \\ \Rightarrow x = 1 \hfill \\$

Here, ${f_1}(x) = {x^2} - 2x + 7$

$\begin{gathered} {f_1}(0) = 0 - 2*0 + 7 \Rightarrow {f_1}(0) = 7 \hfill \\ {f_1}(1) = {1^2} - 2*1 + 7 \Rightarrow {f_1}(1) = 6 \hfill \\ {f_1}( - 3) = {( - 3)^2} - 2( - 3) + 7 \Rightarrow {f_1}(x) = 22 \hfill \\ \end{gathered}$

From the above values, we see that the absolute maximum is 22

and it occur at x = -3.

Here, $f_1'(x)$ is decreasing in [-3, 0] and positive.

Now, differentiate ${f_2}(x)$

$\begin{gathered} \Rightarrow f_2'(x) = \frac{d}{{dx}}{e^{\left( {4{x^3} - 12{x^2} - 180x + 31} \right)}} \hfill \\ \Rightarrow f_2'(x) = {e^{\left( {4{x^3} - 12{x^2} - 180x + 31} \right)}}\frac{d}{{dx}}\left( {4{x^3} - 12{x^2} - 180x + 31} \right) \hfill \\ \Rightarrow f_2'(x) = {e^{\left( {4{x^3} - 12{x^2} - 180x + 31} \right)}}\left( {12{x^2} - 24x - 180} \right) \hfill \\ \Rightarrow f_2'(x) = {e^{\left( {4{x^3} - 12{x^2} - 180x + 31} \right)}}\left[ {12(x - 5)(x + 3)} \right] \hfill \\ \end{gathered}$

For critical points,

$\begin{gathered} f_2'(x) = 0 \hfill \\ \Rightarrow 12(x - 5)(x + 3) = 0 \hfill \\ \Rightarrow (x - 5) = 0\& (x + 3) = 0 \hfill \\ \Rightarrow x = 5\& x = - 3 \hfill \\ \end{gathered}$

Here, ${f_2}(x) = {e^{\left( {4{x^3} - 12{x^2} - 180x + 31} \right)}}$

$\begin{gathered} {f_2}(0) = {e^{\left( {4{{(0)}^3} - 12{{(0)}^2} - 180(0) + 31} \right)}} \Rightarrow {f_2}(0) = 2.90 \hfill \\ {f_2}(5) = {e^{\left( {4{{(5)}^3} - 12{{(5)}^2} - 180(5) + 31} \right)}} \Rightarrow {f_2}(5) = 2.86 \hfill \\ {f_2}( - 3) = {e^{\left( {4{{( - 3)}^3} - 12{{( - 3)}^2} - 180( - 3) + 31} \right)}} \Rightarrow {f_2}( - 3) = 3.08 \hfill \\ \end{gathered}$

The above values show that the absolute maximum is 3.08, and it occurs at x = -3.

Here, ${f_2}(x)$ is decreasing in [-3, 0] and positive.

Step 2 of 2:

Both ${f_1}(x) \text{ and } {f_2}(x)$ decreasing in [-3, 0] and positive.

i.e. f(x) is decreasing in [-3, 0] and positive.

Therefore, Absolute maximum value of f(x) occurs at x = -3

So, $\alpha = - 3$

Final Answer:

Hence, Option (B) is correct.