What is the extreme f(x)=4x^3-36x^2+6 ?

The extreme f(x)=4x^3-36x^2+6 is Maximum(0,6),Minimum(6,-426)

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extreme f(x)=4x^3-36x^2+6

extreme points

f (x) = 4 x^{3} - 36 x^{2} + 6

Maximum (0, 6), Minimum (6, - 426)

Solution steps

f^{'} (x) = 12 x^{2} - 72 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

4 x^{3} - 36 x^{2} + 6 : 6

Maximum (0, 6)

Plug

x = 6

into

4 x^{3} - 36 x^{2} + 6 : - 426

Minimum (6, - 426)

Maximum (0, 6), Minimum (6, - 426)

f (x) = x^{2} + y^{2} - 14 x + 12 y - 13

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

x^{2} + x y + y^{2} - 3 x + 2

f (x) = 25 x + \frac{16}{x}

f (x) = x^{2} - 2 x - 3, 0 \leq x \leq 1

What is the extreme f(x)=4x^3-36x^2+6 ?
The extreme f(x)=4x^3-36x^2+6 is Maximum(0,6),Minimum(6,-426)