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

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Answer 1 · 2016-01-17T07:49:11

$x_1=2.430500874043$ and $y_1=-1.4602879768904$ Maximum Point

$x_2=-1.0971675407097$ and $y_2=-0.002674986072485$ Minimum Point

Determine the derivative of $f(x)$

$f' (x)$
$=((x-2)(x-4)^3*1-x[(x-2)*3(x-4)^2 +(x-4)^3 *1] )/[(x-2)(x-4)^3]^2$

Take the numerator then equate to zero

$((x-2)(x-4)^3*1-x[(x-2)*3(x-4)^2 +(x-4)^3 *1] )=0$

simplify

$(x-2)(x-4)^3-3x(x-2)(x-4)^2-x(x-4)^3=0$

Factoring the common term

$(x-4)^2*[(x-2)(x-4)-3x(x-2)-x(x-4)]=0$

$(x-4)^2*(x^2-6x+8-3x^2+6x-x^2+4x)=0$

$(x-4)^2(-3x^2+4x+8)=0$

The values of x are:

$x=4$ an asymptote

$x_1=(4+sqrt(112))/6=2.430500874043$
Use $x_1$ to obtain $y_1=-1.4602879768904$ Maximum

$x_2=(4-sqrt(112))/6=-1.0971675407097$
Use $x_2$ to obtain $y_2=-0.002674986072485$ # Minimum

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