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

The extreme f(x)=4x^3-5x^2+3 is Maximum(0,3),Minimum(5/6 , 199/108)

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

extreme points

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

Maximum (0, 3), Minimum (\frac{5}{6}, \frac{199}{108})

Solution steps

f^{'} (x) = 12 x^{2} - 10 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{5}{6},

Increasing

: \frac{5}{6} < x < \infty

Plug

x = 0

into

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

Maximum (0, 3)

Plug

x = \frac{5}{6}

into

4 x^{3} - 5 x^{2} + 3 : \frac{199}{108}

Minimum (\frac{5}{6}, \frac{199}{108})

Maximum (0, 3), Minimum (\frac{5}{6}, \frac{199}{108})

P (x, y) = x^{2} + 6 x + 9 - 4 y^{2}

f (x) = x^{3} + 3 x^{2} - 9 x + 7

f (x) = 2 x^{3} + 9 x^{2} - 24 x + 5

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

p (x) = e^{x - 3} + e^{- x}

What is the extreme f(x)=4x^3-5x^2+3 ?
The extreme f(x)=4x^3-5x^2+3 is Maximum(0,3),Minimum(5/6 , 199/108)