What is the extreme f(x)=2x^3-x^2-4x+10,-1<= x<= 0 ?

The extreme f(x)=2x^3-x^2-4x+10,-1<= x<= 0 is Minimum(-1,11),Maximum(-2/3 , 314/27),Minimum(0,10)

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extreme f(x)=2x^3-x^2-4x+10,-1<= x<= 0

extreme points

f (x) = 2 x^{3} - x^{2} - 4 x + 10, - 1 \leq x \leq 0

Minimum (- 1, 11), Maximum (- \frac{2}{3}, \frac{314}{27}), Minimum (0, 10)

Solution steps

Find the critical points:

x = - 1, x = - \frac{2}{3}, x = 0

Domain of

2 x^{3} - x^{2} - 4 x + 10, - 1 \leq x \leq 0 : - 1 \leq x \leq 0

Find intervals:

Increasing

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

Decreasing

: - \frac{2}{3} < x < 0

Plug the extreme point

x = - 1

into

2 x^{3} - x^{2} - 4 x + 10 \Rightarrow y = 11

Minimum (- 1, 11)

Plug the extreme point

x = - \frac{2}{3}

into

2 x^{3} - x^{2} - 4 x + 10 \Rightarrow y = \frac{314}{27}

Maximum (- \frac{2}{3}, \frac{314}{27})

Plug the extreme point

x = 0

into

2 x^{3} - x^{2} - 4 x + 10 \Rightarrow y = 10

Minimum (0, 10)

Minimum (- 1, 11), Maximum (- \frac{2}{3}, \frac{314}{27}), Minimum (0, 10)

f (x) = - \frac{4}{3} x^{\frac{3}{4}}

2 sin (\frac{θ}{2})

f (x) = 4 + 8 x + 128 x^{- 1}

f (x) = 4 x^{5} - 25 x^{4} - 40 x^{3} + 2

x^{3} + x^{2} - 5 x + 1

What is the extreme f(x)=2x^3-x^2-4x+10,-1<= x<= 0 ?
The extreme f(x)=2x^3-x^2-4x+10,-1<= x<= 0 is Minimum(-1,11),Maximum(-2/3 , 314/27),Minimum(0,10)