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

The extreme f(x)=-x^2-x+1,-1<= x<= 1 is Minimum(-1,1),Maximum(-1/2 , 5/4),Minimum(1,-1)

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extreme f(x)=-x^2-x+1,-1<= x<= 1

extreme points

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

Minimum (- 1, 1), Maximum (- \frac{1}{2}, \frac{5}{4}), Minimum (1, - 1)

Solution steps

Find the critical points:

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

Domain of

- x^{2} - x + 1, - 1 \leq x \leq 1 : - 1 \leq x \leq 1

Find intervals:

Increasing

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

Decreasing

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

Plug the extreme point

x = - 1

into

- x^{2} - x + 1 \Rightarrow y = 1

Minimum (- 1, 1)

Plug the extreme point

x = - \frac{1}{2}

into

- x^{2} - x + 1 \Rightarrow y = \frac{5}{4}

Maximum (- \frac{1}{2}, \frac{5}{4})

Plug the extreme point

x = 1

into

- x^{2} - x + 1 \Rightarrow y = - 1

Minimum (1, - 1)

Minimum (- 1, 1), Maximum (- \frac{1}{2}, \frac{5}{4}), Minimum (1, - 1)

f (x) = e^{0.5 x} + 25 6^{- 0.5 x}

f (x) = \frac{ln ( x )}{5 x ^{2}}

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

2800 x - x^{3} + 3 x^{2} + 80 x - 500

f (x) = y = \frac{1}{3} x^{3} - 6 x + 1, - 1 \leq x \leq 6

What is the extreme f(x)=-x^2-x+1,-1<= x<= 1 ?
The extreme f(x)=-x^2-x+1,-1<= x<= 1 is Minimum(-1,1),Maximum(-1/2 , 5/4),Minimum(1,-1)