What is the extreme f(x)=*0.002d^2+0.7d+6.9 ?

The extreme f(x)=*0.002d^2+0.7d+6.9 is Minimum(-175,-54.35)

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extreme f(x)=*0.002d^2+0.7d+6.9

extreme points

f (x) = \cdot 0.002 d^{2} + 0.7 d + 6.9

Minimum (- 175, - 54.35)

Solution steps

f^{'} (d) = 0.004 d + 0.7

Find intervals:

Decreasing

: - \infty < d < - 175,

Increasing

: - 175 < d < \infty

Plug

d = - 175

into

\cdot 0.002 d^{2} + 0.7 d + 6.9 : - 54.35

Minimum (- 175, - 54.35)

f (x) = x + 16 y^{2} - x^{2} + x y - 3 y^{2} + 25

f (x, y) = 3 x^{2} + 3 y^{2} - 6 x + 12 y

\frac{3 x}{x ^{2} - 1}

f (x) = \frac{9 ( x ^{2} - 3 )}{x ^{3}}

y = x^{4} - 2 x^{2} + 1

What is the extreme f(x)=*0.002d^2+0.7d+6.9 ?
The extreme f(x)=*0.002d^2+0.7d+6.9 is Minimum(-175,-54.35)