Question

Let f(x)=13x3−x2−3x+10 defined on the interval [−4,4].Let x1 and x2 be the critical points of f, and let's impose that x1

Answer 1

Answer:

$f(x_1)=2\sqrt{118}\\f(x_2)=-2\sqrt{118}$

Step-by-step explanation:

$f(x)=13x^3-x^2-3x+10$

Differentiate with respect to $x$

$f'(x)=39x^2-2x-3$

For equation of the form $ax^2+bx+c=0$, solutions are given by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

To find: critical points

$39x^2-2x-3=0$

$x=\frac{2\pm \sqrt{4+468}}{78}\\=\frac{2\pm \sqrt{472}}{78}\\=\frac{2\pm 2\sqrt{118}}{78}\\=\frac{1\pm \sqrt{118}}{39}$

Let $x_1=\frac{1+\sqrt{118}}{39}\,,\,x_2=\frac{1- \sqrt{118}}{39}$

Differentiate $f'(x)$ again with respect to $x$

$f''(x)=78x-2$

$f(x_1)=f\left (\frac{1+\sqrt{118}}{39} \right )\\=78\left ( \frac{1+\sqrt{118}}{39} \right )-2\\=2\left ( 1+\sqrt{118} \right )-2\\=2\sqrt{118}\\f(x_2)=f\left (\frac{1-\sqrt{118}}{39} \right )\\=78\left ( \frac{1-\sqrt{118}}{39} \right )-2\\=2\left ( 1-\sqrt{118} \right )-2\\=-2\sqrt{118}$