What is the extreme f(x)=x^3-2x^2-15x+10,-2<= x<= 0 ?

The extreme f(x)=x^3-2x^2-15x+10,-2<= x<= 0 is Minimum(-2,24),Maximum(-5/3 , 670/27),Minimum(0,10)

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extreme f(x)=x^3-2x^2-15x+10,-2<= x<= 0

extreme points

f (x) = x^{3} - 2 x^{2} - 15 x + 10, - 2 \leq x \leq 0

Minimum (- 2, 24), Maximum (- \frac{5}{3}, \frac{670}{27}), Minimum (0, 10)

Solution steps

Find the critical points:

x = - 2, x = - \frac{5}{3}, x = 0

Domain of

x^{3} - 2 x^{2} - 15 x + 10, - 2 \leq x \leq 0 : - 2 \leq x \leq 0

Find intervals:

Increasing

: - 2 < x < - \frac{5}{3},

Decreasing

: - \frac{5}{3} < x < 0

Plug the extreme point

x = - 2

into

x^{3} - 2 x^{2} - 15 x + 10 \Rightarrow y = 24

Minimum (- 2, 24)

Plug the extreme point

x = - \frac{5}{3}

into

x^{3} - 2 x^{2} - 15 x + 10 \Rightarrow y = \frac{670}{27}

Maximum (- \frac{5}{3}, \frac{670}{27})

Plug the extreme point

x = 0

into

x^{3} - 2 x^{2} - 15 x + 10 \Rightarrow y = 10

Minimum (0, 10)

Minimum (- 2, 24), Maximum (- \frac{5}{3}, \frac{670}{27}), Minimum (0, 10)

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

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

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

f (x) = \frac{1}{16 - x ^{2} + 4 y ^{2}}

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

What is the extreme f(x)=x^3-2x^2-15x+10,-2<= x<= 0 ?
The extreme f(x)=x^3-2x^2-15x+10,-2<= x<= 0 is Minimum(-2,24),Maximum(-5/3 , 670/27),Minimum(0,10)