What is the extreme f(x)= 49/2 x^2-ln(x) ?

The extreme f(x)= 49/2 x^2-ln(x) is Minimum(1/7 , 1/2+ln(7))

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extreme f(x)= 49/2 x^2-ln(x)

extreme points

f (x) = \frac{49}{2} x^{2} - ln (x)

Minimum (\frac{1}{7}, \frac{1}{2} + ln (7))

Solution steps

f^{'} (x) = 49 x - \frac{1}{x}

Find intervals:

Decreasing

: 0 < x < \frac{1}{7},

Increasing

: \frac{1}{7} < x < \infty

Plug

x = \frac{1}{7}

into

\frac{49}{2} x^{2} - ln (x) : \frac{1}{2} + ln (7)

Minimum (\frac{1}{7}, \frac{1}{2} + ln (7))

f (x) = 4 x^{3} e^{- x}, - 1 \leq x \leq 4

x^{4} - 4 x^{2} + 2

(x^{2} - 4)^{7}

f (x, y) = \frac{1}{16 - 4 x ^{2} - y ^{2}}

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

What is the extreme f(x)= 49/2 x^2-ln(x) ?
The extreme f(x)= 49/2 x^2-ln(x) is Minimum(1/7 , 1/2+ln(7))