What is the extreme f(x)=4x^3-x^2+9 ?

The extreme f(x)=4x^3-x^2+9 is Maximum(0,9),Minimum(1/6 , 971/108)

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extreme f(x)=4x^3-x^2+9

extreme

Maximum (0, 9), Minimum (\frac{1}{6}, \frac{971}{108})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{1}{6},

Increasing

: \frac{1}{6} < x < \infty

Plug

x = 0

into

4 x^{3} - x^{2} + 9 : 9

Maximum (0, 9)

Plug

x = \frac{1}{6}

into

4 x^{3} - x^{2} + 9 : \frac{971}{108}

Minimum (\frac{1}{6}, \frac{971}{108})

Maximum (0, 9), Minimum (\frac{1}{6}, \frac{971}{108})

What is the extreme f(x)=4x^3-x^2+9 ?
The extreme f(x)=4x^3-x^2+9 is Maximum(0,9),Minimum(1/6 , 971/108)