What is the extreme (3x-2)/(sqrt(2x^2+1)) ?

The extreme (3x-2)/(sqrt(2x^2+1)) is Minimum(-3/4 ,-sqrt(17/2))

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extreme (3x-2)/(sqrt(2x^2+1))

extreme points

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

Minimum (- \frac{3}{4}, - \frac{17}{2})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - \frac{3}{4},

Increasing

: - \frac{3}{4} < x < \infty

Plug

x = - \frac{3}{4}

into

\frac{3 x - 2}{2 x ^{2} + 1} : - \frac{17}{2}

Minimum (- \frac{3}{4}, - \frac{17}{2})

f (x) = 12 x^{2} - 16

2 x^{3} - 6 x^{2} - 210 x + 7

f (x, y) = 2 x^{2} - 8 y^{2} - 13

f (x) = 2 x - 5 x^{\frac{2}{5}}

y = - x^{2} + 5, (- 2, 4)

What is the extreme (3x-2)/(sqrt(2x^2+1)) ?
The extreme (3x-2)/(sqrt(2x^2+1)) is Minimum(-3/4 ,-sqrt(17/2))