What is the extreme f(x)=sqrt((x-1)^2+(4-4x^2)) ?

The extreme f(x)=sqrt((x-1)^2+(4-4x^2)) is Minimum(-5/3 ,0),Maximum(-1/3 , 4/(sqrt(3))),Minimum(1,0)

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extreme f(x)=sqrt((x-1)^2+(4-4x^2))

extreme points

f (x) = (x - 1)^{2} + (4 - 4 x^{2})

Minimum (- \frac{5}{3}, 0), Maximum (- \frac{1}{3}, \frac{4}{3}), Minimum (1, 0)

Solution steps

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

Find intervals:

Increasing

: - \frac{5}{3} < x < - \frac{1}{3},

Decreasing

: - \frac{1}{3} < x < 1

Plug

x = - \frac{5}{3}

into

(x - 1)^{2} + 4 - 4 x^{2} : 0

Minimum (- \frac{5}{3}, 0)

Plug

x = - \frac{1}{3}

into

(x - 1)^{2} + 4 - 4 x^{2} : \frac{4}{3}

Maximum (- \frac{1}{3}, \frac{4}{3})

Plug

x = 1

into

(x - 1)^{2} + 4 - 4 x^{2} : 0

Minimum (1, 0)

Minimum (- \frac{5}{3}, 0), Maximum (- \frac{1}{3}, \frac{4}{3}), Minimum (1, 0)

f (x) = x^{3} + y^{3} - 30 x y

f (x) = 3 x^{4} - 54 x^{2} + 4

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

f (x) = x^{3} - 2 x^{2} - 4 x

y = \frac{x ^{3} ( 3 x - 8 )}{12}

What is the extreme f(x)=sqrt((x-1)^2+(4-4x^2)) ?
The extreme f(x)=sqrt((x-1)^2+(4-4x^2)) is Minimum(-5/3 ,0),Maximum(-1/3 , 4/(sqrt(3))),Minimum(1,0)