What is the critical f(x)=sqrt(x^2-3x+6) ?

The critical f(x)=sqrt(x^2-3x+6) is x= 3/2

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critical f(x)=sqrt(x^2-3x+6)

critical points

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

x = \frac{3}{2}

Solution steps

x^{2} - 3 x + 6

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{3}{2}

Identify critical points not in

f (x)

domain

Domain of

x^{2} - 3 x + 6 : - \infty < x < \infty

All critical points are in domain

x = \frac{3}{2}

f (x, y) = x^{8} + 2 y^{6}

f (x) = x^{3} - 6 x^{2} + 9

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

2 x^{3} - 6 x + 4

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

What is the critical f(x)=sqrt(x^2-3x+6) ?
The critical f(x)=sqrt(x^2-3x+6) is x= 3/2