# What are the local extrema of f(x)= x/((x-2)(x-4)^3)?

#### Answer:

${x}_{1} = 2.430500874043$ and ${y}_{1} = - 1.4602879768904$ Maximum Point

${x}_{2} = - 1.0971675407097$ and ${y}_{2} = - 0.002674986072485$ Minimum Point

#### Explanation:

Determine the derivative of $f \left(x\right)$

$f ' \left(x\right)$
$= \frac{\left(x - 2\right) {\left(x - 4\right)}^{3} \cdot 1 - x \left[\left(x - 2\right) \cdot 3 {\left(x - 4\right)}^{2} + {\left(x - 4\right)}^{3} \cdot 1\right]}{\left(x - 2\right) {\left(x - 4\right)}^{3}} ^ 2$

Take the numerator then equate to zero

$\left(\left(x - 2\right) {\left(x - 4\right)}^{3} \cdot 1 - x \left[\left(x - 2\right) \cdot 3 {\left(x - 4\right)}^{2} + {\left(x - 4\right)}^{3} \cdot 1\right]\right) = 0$

simplify

$\left(x - 2\right) {\left(x - 4\right)}^{3} - 3 x \left(x - 2\right) {\left(x - 4\right)}^{2} - x {\left(x - 4\right)}^{3} = 0$

Factoring the common term

${\left(x - 4\right)}^{2} \cdot \left[\left(x - 2\right) \left(x - 4\right) - 3 x \left(x - 2\right) - x \left(x - 4\right)\right] = 0$

${\left(x - 4\right)}^{2} \cdot \left({x}^{2} - 6 x + 8 - 3 {x}^{2} + 6 x - {x}^{2} + 4 x\right) = 0$

${\left(x - 4\right)}^{2} \left(- 3 {x}^{2} + 4 x + 8\right) = 0$

The values of x are:

$x = 4$ an asymptote

${x}_{1} = \frac{4 + \sqrt{112}}{6} = 2.430500874043$
Use ${x}_{1}$ to obtain ${y}_{1} = - 1.4602879768904$ Maximum

${x}_{2} = \frac{4 - \sqrt{112}}{6} = - 1.0971675407097$
Use ${x}_{2}$ to obtain ${y}_{2} = - 0.002674986072485$# Minimum