What is the critical f(w)=(w^2+2w+1)/(3w-5) ?

The critical f(w)=(w^2+2w+1)/(3w-5) is w=-1,w= 13/3

critical f(w)=(w^2+2w+1)/(3w-5)

critical points

f (w) = \frac{w ^{2} + 2 w + 1}{3 w - 5}

w = - 1, w = \frac{13}{3}

Solution steps

\frac{w ^{2} + 2 w + 1}{3 w - 5}

Find where

f^{'} (w)

is equal to zero or undefined

w = - 1, w = \frac{13}{3}, w = \frac{5}{3}

Identify critical points not in

f (w)

domain

Domain of

\frac{w ^{2} + 2 w + 1}{3 w - 5} : w < \frac{5}{3} or w > \frac{5}{3}

\frac{w ^{2} + 2 w + 1}{3 w - 5}

is not defined at

w = \frac{5}{3}

, therefore:

w = - 1, w = \frac{13}{3}

f (x) = x^{3} (x + 2)

f (x) = 2 x^{3} - 3 x^{2} y - 3 y^{2}

f (x) = \frac{x - 3}{x ^{2} - 5 x + 6}

\frac{1}{2} x^{4} - 3 x y + 3 y^{4}

f (x) = x^{3} + 3 x^{2} + 3 x + 1

What is the critical f(w)=(w^2+2w+1)/(3w-5) ?
The critical f(w)=(w^2+2w+1)/(3w-5) is w=-1,w= 13/3