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

The extreme f(x)=-3x^2-x+2,-4<= x<= 4 is Minimum(-4,-42),Maximum(-1/6 , 25/12),Minimum(4,-50)

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

extreme points

f (x) = - 3 x^{2} - x + 2, - 4 \leq x \leq 4

Minimum (- 4, - 42), Maximum (- \frac{1}{6}, \frac{25}{12}), Minimum (4, - 50)

Solution steps

Find the critical points:

x = - 4, x = - \frac{1}{6}, x = 4

Domain of

- 3 x^{2} - x + 2, - 4 \leq x \leq 4 : - 4 \leq x \leq 4

Find intervals:

Increasing

: - 4 < x < - \frac{1}{6},

Decreasing

: - \frac{1}{6} < x < 4

Plug the extreme point

x = - 4

into

- 3 x^{2} - x + 2 \Rightarrow y = - 42

Minimum (- 4, - 42)

Plug the extreme point

x = - \frac{1}{6}

into

- 3 x^{2} - x + 2 \Rightarrow y = \frac{25}{12}

Maximum (- \frac{1}{6}, \frac{25}{12})

Plug the extreme point

x = 4

into

- 3 x^{2} - x + 2 \Rightarrow y = - 50

Minimum (4, - 50)

Minimum (- 4, - 42), Maximum (- \frac{1}{6}, \frac{25}{12}), Minimum (4, - 50)

\frac{7 + 2 x}{x}

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f (x) = 99 - \frac{x}{20}

f (x) = x (20 - 2 x) (\frac{48}{2} - x)

f (x, y) = 6 x^{2} y - 9 x y^{3} + 3 y^{3}

What is the extreme f(x)=-3x^2-x+2,-4<= x<= 4 ?
The extreme f(x)=-3x^2-x+2,-4<= x<= 4 is Minimum(-4,-42),Maximum(-1/6 , 25/12),Minimum(4,-50)