Answer
a) the maximum value will be 30 at the point $x=1$.
b) the minimum value will be 3 at $x=4$.
Work Step by Step
(a)
Let us consider the function $2{{x}^{3}}-15{{x}^{2}}+24x+19$.
In order to calculate the local maxima or minima, differentiate the function with respect to x and put it equal to zero as follows:
$\begin{align}
& \frac{\text{d}\left( 2{{x}^{3}}-15{{x}^{2}}+24x+19 \right)}{\text{d}x}=0 \\
& 2\left( 3{{x}^{2}} \right)-15\left( 2x \right)+24+0=0 \\
& 6{{x}^{2}}-30x+24=0 \\
& {{x}^{2}}-5x+4=0
\end{align}$
Now simplify the above equation further to find the zeros:
$\begin{align}
& {{x}^{2}}-4x-x+4=0 \\
& x\left( x-4 \right)-1\left( x-4 \right)=0 \\
& \left( x-4 \right)\left( x-1 \right)=0 \\
& x=4\text{ and }x=1
\end{align}$
Substitute $x=4\text{ and }x=1$ one by one in the main equation
For $x=4$
$\begin{align}
& 2{{\left( 4 \right)}^{3}}-15{{\left( 4 \right)}^{2}}+24\left( 4 \right)+19=128-240+96+19 \\
& =3
\end{align}$
And for $x=1$
$\begin{align}
& 2{{\left( 1 \right)}^{3}}-15{{\left( 1 \right)}^{2}}+24\left( 1 \right)+19=2-15+24+19 \\
& =30
\end{align}$
Thus, the higher value among the two values is 30. So, the maximum value will be 30 at the point $x=1$.
(b)
Let us consider the function $2{{x}^{3}}-15{{x}^{2}}+24x+19$.
In order to calculate the local maxima or minima, differentiate the function with respect to x as follows:
$\begin{align}
& \frac{\text{d}\left( 2{{x}^{3}}-15{{x}^{2}}+24x+19 \right)}{\text{d}x}=0 \\
& 2\left( 3{{x}^{2}} \right)-15\left( 2x \right)+24+0=0 \\
& 6{{x}^{2}}-30x+24=0 \\
& {{x}^{2}}-5x+4=0
\end{align}$
Now simplify the equation further and get the zeros as follows:
$\begin{align}
& {{x}^{2}}-4x-x+4=0 \\
& x\left( x-4 \right)-1\left( x-4 \right)=0 \\
& \left( x-4 \right)\left( x-1 \right)=0 \\
& x=4\text{ and }x=1
\end{align}$
Substitute $x=4\text{ and }x=1$ one by one in the main equation
For $x=4$
$\begin{align}
& 2{{\left( 4 \right)}^{3}}-15{{\left( 4 \right)}^{2}}+24\left( 4 \right)+19=128-240+96+19 \\
& =3
\end{align}$
And for $x=1$
$\begin{align}
& 2{{\left( 1 \right)}^{3}}-15{{\left( 1 \right)}^{2}}+24\left( 1 \right)+19=2-15+24+19 \\
& =30
\end{align}$
Thus, the lower value among the two values is 3.
