What is the critical f(x)=2.6+2.2x-0.6x^2 ?

The critical f(x)=2.6+2.2x-0.6x^2 is x= 11/6

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critical f(x)=2.6+2.2x-0.6x^2

critical points

f (x) = 2.6 + 2.2 x - 0.6 x^{2}

x = \frac{11}{6}

Solution steps

2.6 + 2.2 x - 0.6 x^{2}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{11}{6}

Identify critical points not in

f (x)

domain

Domain of

2.6 + 2.2 x - 0.6 x^{2} : - \infty < x < \infty

All critical points are in domain

x = \frac{11}{6}

f (x) = ln (\frac{x ^{2} - 3}{1 - x ^{2}})

2 x + 2 y = 4

f (x) = 8 x, x \geq 0

f (x) = (\frac{1}{5})

4 x - 1 = 3 y + 5

What is the critical f(x)=2.6+2.2x-0.6x^2 ?
The critical f(x)=2.6+2.2x-0.6x^2 is x= 11/6