Question

Find the maximum value of ${2x}^3-24x+107$ in the interval $[1,3]$.

Answer 1

$f\left(x\right)={2x}^3-24x+107$

Differentiating with respect to x.

$f\text{'}\left(x\right)={6x}^2-24$

For maxima and minima

$f\text{'}\left(x\right)=0$

${6x}^2-24=0$

${6x}^2=24$

$x^2=\frac{24}{6}$

$x^2=4$

$x=\pm 2$

$f\left(2\right)=2{\left(2\right)}^3-24\times 2+107$

$=16-48+107$

$=75$