What is the extreme f(x)=-2x^2+x-1,(-3,5) ?

The extreme f(x)=-2x^2+x-1,(-3,5) is Maximum(1/4 ,-7/8)

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extreme f(x)=-2x^2+x-1,(-3,5)

extreme points

f (x) = - 2 x^{2} + x - 1, (- 3, 5)

Maximum (\frac{1}{4}, - \frac{7}{8})

Solution steps

Find the critical points:

x = \frac{1}{4}

Domain of

- 2 x^{2} + x - 1, (- 3, 5) : - 3 < x < 5

Find intervals:

Increasing

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

Decreasing

: \frac{1}{4} < x < 5

Plug the extreme point

x = \frac{1}{4}

into

- 2 x^{2} + x - 1 \Rightarrow y = - \frac{7}{8}

Maximum (\frac{1}{4}, - \frac{7}{8})

f (x) = 6 x^{2} - 48 x + 72

y = - 0.05 x^{2} + 50 x - 600

g (x) = - 4 x^{3} - 66 x^{2} - 360 x + 7

f (x) = 4 sin (x) cos (x) + 12

x (57 - x^{2})

What is the extreme f(x)=-2x^2+x-1,(-3,5) ?
The extreme f(x)=-2x^2+x-1,(-3,5) is Maximum(1/4 ,-7/8)