Answer
a) the maximum value will be 21 at the point $x=-2$.
b) the minimum value will be 21 at $x=1$.
Work Step by Step
(a)
Let us consider the function $2{{x}^{3}}+3{{x}^{2}}-12x+1$.
In order to calculate the local maxima or minima, differentiate the function with respect to x and put it equal to zero as follows:
$\begin{align}
& \frac{\text{d}\left( 2{{x}^{3}}+3{{x}^{2}}-12x+1 \right)}{\text{d}x}=0 \\
& 2\left( 3{{x}^{2}} \right)+3\left( 2x \right)-12+0=0 \\
& 6{{x}^{2}}+6x-12=0 \\
& {{x}^{2}}+x-2=0
\end{align}$
Now simplify the above equation further to find the zeros:
$\begin{align}
& {{x}^{2}}+2x-x-2=0 \\
& x\left( x+2 \right)-1\left( x+2 \right)=0 \\
& \left( x+2 \right)\left( x-1 \right)=0 \\
& x=-2\text{ and }x=1
\end{align}$
Substitute $x=-2\text{ and }x=1$ one by one in the main equation.
For $x=-2$
$\begin{align}
& 2{{\left( -2 \right)}^{3}}+3{{\left( -2 \right)}^{2}}-12\left( -2 \right)+1=-16+12+24+1 \\
& =21
\end{align}$
And for $x=1$
$\begin{align}
& 2{{\left( 1 \right)}^{3}}+3{{\left( 1 \right)}^{2}}-12\left( 1 \right)+1=2+3-12+1 \\
& =-6
\end{align}$
Thus, the higher value among the two values is 21. So, the maximum value will be 21 at the point $x=-2$.
(b)
Let us consider the function $2{{x}^{3}}+3{{x}^{2}}-12x+1$.
In order to calculate the local maxima or minima, differentiate the function with respect to x as follows:
$\begin{align}
& \frac{\text{d}\left( 2{{x}^{3}}+3{{x}^{2}}-12x+1 \right)}{\text{d}x}=0 \\
& 2\left( 3{{x}^{2}} \right)+3\left( 2x \right)-12+0=0 \\
& 6{{x}^{2}}+6x-12=0 \\
& {{x}^{2}}+x-2=0
\end{align}$
Now simplify the equation further and get the zeros as follows:
$\begin{align}
& {{x}^{2}}+2x-x-2=0 \\
& x\left( x+2 \right)-1\left( x+2 \right)=0 \\
& \left( x+2 \right)\left( x-1 \right)=0 \\
& x=-2\text{ and }x=1
\end{align}$
Substitute $x=-2\text{ and }x=1$ one by one in the main equation.
For $x=-2$
$\begin{align}
& 2{{\left( -2 \right)}^{3}}+3{{\left( -2 \right)}^{2}}-12\left( -2 \right)+1=-16+12+24+1 \\
& =21
\end{align}$
And for $x=1$
$\begin{align}
& 2{{\left( 1 \right)}^{3}}+3{{\left( 1 \right)}^{2}}-12\left( 1 \right)+1=2+3-12+1 \\
& =-6
\end{align}$
Thus, the smaller value among the two values is -6. So, the minimum value will be 21 at $x=1$.
