What is the extreme f(x)=-4x^3-3x^2+6x+1 ?

The extreme f(x)=-4x^3-3x^2+6x+1 is Minimum(-1,-4),Maximum(1/2 , 11/4)

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extreme f(x)=-4x^3-3x^2+6x+1

extreme points

f (x) = - 4 x^{3} - 3 x^{2} + 6 x + 1

Minimum (- 1, - 4), Maximum (\frac{1}{2}, \frac{11}{4})

Solution steps

f^{'} (x) = - 12 x^{2} - 6 x + 6

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < \infty

Plug

x = - 1

into

- 4 x^{3} - 3 x^{2} + 6 x + 1 : - 4

Minimum (- 1, - 4)

Plug

x = \frac{1}{2}

into

- 4 x^{3} - 3 x^{2} + 6 x + 1 : \frac{11}{4}

Maximum (\frac{1}{2}, \frac{11}{4})

Minimum (- 1, - 4), Maximum (\frac{1}{2}, \frac{11}{4})

150 (x y)^{2} + (0.5 x + 2 y - 2)^{2}

3 x^{2} - 5 x - 1

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

y = \frac{( 2 x + 3 )}{x + 2}

f (x) = \frac{x ^{2} - 5 x + 7}{x - 3}

What is the extreme f(x)=-4x^3-3x^2+6x+1 ?
The extreme f(x)=-4x^3-3x^2+6x+1 is Minimum(-1,-4),Maximum(1/2 , 11/4)