What is the extreme f(x)= 7/2 x^2+7/2-7x-4 ?

The extreme f(x)= 7/2 x^2+7/2-7x-4 is Minimum(1,-4)

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extreme f(x)= 7/2 x^2+7/2-7x-4

extreme points

f (x) = \frac{7}{2} x^{2} + \frac{7}{2} - 7 x - 4

Minimum (1, - 4)

Solution steps

f^{'} (x) = 7 x - 7

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 1

into

\frac{7}{2} x^{2} + \frac{7}{2} - 7 x - 4 : - 4

Minimum (1, - 4)

\frac{2 ln ( x )}{x}

f (x) = x^{4} - 24 x^{2} + 64 x + 16

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

f (x) = \frac{x ^{3}}{e ^{8 x}}

f (x) = x e^{- \frac{x ^{2}}{32}}

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