What is the extreme f(x)=4x^3-4x^2-4x+5 ?

The extreme f(x)=4x^3-4x^2-4x+5 is Maximum(-1/3 , 155/27),Minimum(1,1)

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extreme f(x)=4x^3-4x^2-4x+5

extreme

Maximum (- \frac{1}{3}, \frac{155}{27}), Minimum (1, 1)

Solution steps

f^{'} (x) = 12 x^{2} - 8 x - 4

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 1 < x < \infty

Plug

x = - \frac{1}{3}

into

4 x^{3} - 4 x^{2} - 4 x + 5 : \frac{155}{27}

Maximum (- \frac{1}{3}, \frac{155}{27})

Plug

x = 1

into

4 x^{3} - 4 x^{2} - 4 x + 5 : 1

Minimum (1, 1)

Maximum (- \frac{1}{3}, \frac{155}{27}), Minimum (1, 1)

What is the extreme f(x)=4x^3-4x^2-4x+5 ?
The extreme f(x)=4x^3-4x^2-4x+5 is Maximum(-1/3 , 155/27),Minimum(1,1)