What is the extreme y=5x^2ln(x/4) ?

The extreme y=5x^2ln(x/4) is Minimum(4/(sqrt(e)),-(40)/e)

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extreme y=5x^2ln(x/4)

extreme points

y = 5 x^{2} ln (\frac{x}{4})

Minimum (\frac{4}{e}, - \frac{40}{e})

Solution steps

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

Find intervals:

Decreasing

: 0 < x < \frac{4}{e},

Increasing

: \frac{4}{e} < x < \infty

Plug

x = \frac{4}{e}

into

5 x^{2} ln (\frac{x}{4}) : - \frac{40}{e}

Minimum (\frac{4}{e}, - \frac{40}{e})

\frac{x ^{2} - 8}{x + 3}

f (x) = \frac{- 5}{x - 6}

f (x, y) = 400 - 9 x^{2} - 64 y^{2}

f (x) = x^{3} - 6 x^{2} - 15 x + 2

9 x ln (x)

What is the extreme y=5x^2ln(x/4) ?
The extreme y=5x^2ln(x/4) is Minimum(4/(sqrt(e)),-(40)/e)