What is the extreme f(x)=sqrt(82x^2+180x+100) ?

The extreme f(x)=sqrt(82x^2+180x+100) is Minimum(-45/41 ,(5sqrt(2))/(sqrt(41)))

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extreme f(x)=sqrt(82x^2+180x+100)

extreme points

f (x) = 82 x^{2} + 180 x + 100

Minimum (- \frac{45}{41}, \frac{5 2}{41})

Solution steps

f^{'} (x) = \frac{2 ( 41 x + 45 )}{82 x ^{2} + 180 x + 100}

Find intervals:

Decreasing

: - \infty < x < - \frac{45}{41},

Increasing

: - \frac{45}{41} < x < \infty

Plug

x = - \frac{45}{41}

into

82 x^{2} + 180 x + 100 : \frac{5 2}{41}

Minimum (- \frac{45}{41}, \frac{5 2}{41})

P (x, y) = x^{2} + y^{2} + 16 x - 16 y

f (x) = - 0.001 x^{2} + 9 x - 1296

f (x) = \frac{( 500 + 500 \cdot x )}{1 + 0.04 \cdot x ^{2}}

\frac{( x ^{2} + 2 )}{( x ^{2} - 16 )}

y = 2 x^{2} + 5

What is the extreme f(x)=sqrt(82x^2+180x+100) ?
The extreme f(x)=sqrt(82x^2+180x+100) is Minimum(-45/41 ,(5sqrt(2))/(sqrt(41)))