# What are the local extrema of f(x)=(x-3)(x^2-2x-5)?

Jan 14, 2018

$f \left(x\right)$ has a local maximum at $\approx \left(0.1032 , 15.0510\right)$
$f \left(x\right)$ has a local minimum at $\approx \left(3.2301 , - 0.2362\right)$

#### Explanation:

$f \left(x\right) = \left(x - 3\right) \left({x}^{2} - 2 x - 5\right)$

Apply product rule.

f'(x) = (x-3) * d/dx (x^2-2x-5) + d/dx(x-3)* (x^2-2x-5)

Apply power rule.

$f ' \left(x\right) = \left(x - 3\right) \left(2 x - 2\right) + 1 \cdot \left({x}^{2} - 2 x - 5\right)$

$= 2 {x}^{2} - 8 x + 6 + {x}^{2} - 2 x - 5$

$= 3 {x}^{2} - 10 x + 1$

For local extrema $f ' \left(x\right) = 0$

Hence, $3 {x}^{2} - 10 x + 1 = 0$

$x = \frac{+ 10 \pm \sqrt{{\left(- 10\right)}^{2} - 4 \cdot 3 \cdot 1}}{2 \cdot 3}$

$= \frac{10 \pm \sqrt{88}}{6}$

$\approx 3.2301 \mathmr{and} 0.1032$

$f ' ' \left(x\right) = 6 x - 10$

For local maximum $f ' ' < 0$ at extreme point.

For local minimum $f ' ' > 0$ at extreme point.

Testing $f ' ' \left(3.2301\right) > 0 \to f \left(3.2301\right) = {f}_{\min}$

Testing $f ' ' \left(0.1032\right) < 0 \to f \left(0.1032\right) = {f}_{\max}$

Hence, ${f}_{\max} \approx \left(0.1032 - 3\right) \left({0.1032}^{2} - 2 \cdot 0.1032 - 5\right)$

$\approx 15.0510$

And, ${f}_{\min} \approx \left(3.2301 - 3\right) \left({3.2301}^{2} - 2 \cdot 3.2301 - 5\right)$

$\approx - 0.2362$

$\therefore f \left(x\right)$ has a local maximum at $\approx \left(0.1032 , 15.0510\right)$

$\mathmr{and} f \left(x\right)$ has a local minimum at $\approx \left(3.2301 , - 0.2362\right)$

We can see these local extrema by zooming to the relevant points on the graph of $f \left(x\right)$ below.

graph{(x-3)(x^2-2x-5) [-29.02, 28.72, -6.2, 22.63]}