What is the extreme f(x)=-7(x+1)^2+3x+7 ?

The extreme f(x)=-7(x+1)^2+3x+7 is Maximum(-11/14 , 121/28)

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extreme f(x)=-7(x+1)^2+3x+7

extreme points

f (x) = - 7 (x + 1)^{2} + 3 x + 7

Maximum (- \frac{11}{14}, \frac{121}{28})

Solution steps

f^{'} (x) = - 14 x - 11

Find intervals:

Increasing

: - \infty < x < - \frac{11}{14},

Decreasing

: - \frac{11}{14} < x < \infty

Plug

x = - \frac{11}{14}

into

- 7 (x + 1)^{2} + 3 x + 7 : \frac{121}{28}

Maximum (- \frac{11}{14}, \frac{121}{28})

y = 2 x^{3} - 3 x + 11

y = - x^{2} + 1, - 2 \leq x \leq 1

f (x) = x^{3} - 6 x^{2} - 15 x

f (x) = \frac{( x ^{2} - 3 )}{x - 2}

f (x) = 4 x + \frac{9}{x}

What is the extreme f(x)=-7(x+1)^2+3x+7 ?
The extreme f(x)=-7(x+1)^2+3x+7 is Maximum(-11/14 , 121/28)