## Precalculus (6th Edition) Blitzer

a) the maximum value will be 30 at the point $x=1$. b) the minimum value will be 3 at $x=4$.
(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.