What is the extreme f(x)=(1+3x)/(sqrt(4+5x^2)) ?

The extreme f(x)=(1+3x)/(sqrt(4+5x^2)) is Maximum(12/5 ,(sqrt(41))/(2sqrt(5)))

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extreme f(x)=(1+3x)/(sqrt(4+5x^2))

extreme points

f (x) = \frac{1 + 3 x}{4 + 5 x ^{2}}

Maximum (\frac{12}{5}, \frac{41}{2 5})

Solution steps

f^{'} (x) = \frac{- 5 x + 12}{( 4 + 5 x ^{2} ) 4 + 5 x ^{2}}

Find intervals:

Increasing

: - \infty < x < \frac{12}{5},

Decreasing

: \frac{12}{5} < x < \infty

Plug

x = \frac{12}{5}

into

\frac{1 + 3 x}{4 + 5 x ^{2}} : \frac{41}{2 5}

Maximum (\frac{12}{5}, \frac{41}{2 5})

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

f (x) = (2 x^{2} + 1)^{2}

f (x, y) = \frac{100}{1 + x ^{2} + y ^{2}}

f (x) = \frac{1}{6} x^{3} - x^{2} + 6 x + 11

f (x, y) = x^{2} + y^{2} + 8 x - 2 y

What is the extreme f(x)=(1+3x)/(sqrt(4+5x^2)) ?
The extreme f(x)=(1+3x)/(sqrt(4+5x^2)) is Maximum(12/5 ,(sqrt(41))/(2sqrt(5)))