How do you find the relative extrema for f(x)=x^3-4x^2+x+6?

Jul 21, 2018

relative min at $\left(\frac{4}{3} + \frac{\sqrt{13}}{3} , \frac{70}{27} - \frac{26 \sqrt{13}}{27}\right)$

relative max at $\left(\frac{4}{3} - \frac{\sqrt{13}}{3} , \frac{70}{27} + \frac{26 \sqrt{13}}{27}\right)$

Explanation:

Given: $f \left(x\right) = {x}^{3} - 4 {x}^{2} + x + 6$

To find relative extrema first find the first derivative :

$f ' \left(x\right) = 3 {x}^{2} - 8 x + 1$

Find the critical value(s) by setting $f ' \left(x\right) = 0$

$f ' \left(x\right) = 3 {x}^{2} - 8 x + 1 = 0$

Use the quadratic formula to find the critical value(s):

$x = \frac{8 \pm \sqrt{{8}^{2} - 4 \left(3\right) \left(1\right)}}{2 \cdot 3} = \frac{4}{3} \pm \frac{\sqrt{52}}{6} = \frac{4}{3} \pm \frac{\sqrt{13}}{3}$

Find critical points :

$f \left(\frac{4}{3} + \frac{\sqrt{13}}{3}\right) = \frac{70}{27} - \frac{26 \sqrt{13}}{27} \approx - .8794$

$f \left(\frac{4}{3} - \frac{\sqrt{13}}{3}\right) = \frac{70}{27} + \frac{26 \sqrt{13}}{27} \approx 6.0646$

Use 2nd derivative test since it is easy to find the 2nd derivative of this function:

$f ' ' \left(x\right) = 6 x - 8$

If $f ' ' \left(c\right) > 0$ we have a relative minimum

If $f ' ' \left(c\right) < 0$ we have a relative maximum

$f ' ' \left(\frac{4}{3} + \frac{\sqrt{13}}{3}\right) > 0$ relative min at $x = \frac{4}{3} + \frac{\sqrt{13}}{3}$

$f ' ' \left(\frac{4}{3} - \frac{\sqrt{13}}{3}\right) < 0$ relative max at $x = \frac{4}{3} - \frac{\sqrt{13}}{3}$