What is the extreme f(x)=7x^2ln(x/4) ?

The extreme f(x)=7x^2ln(x/4) is Minimum(4/(sqrt(e)),-(56)/e)

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extreme f(x)=7x^2ln(x/4)

extreme points

f (x) = 7 x^{2} ln (\frac{x}{4})

Minimum (\frac{4}{e}, - \frac{56}{e})

Solution steps

f^{'} (x) = 7 (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

7 x^{2} ln (\frac{x}{4}) : - \frac{56}{e}

Minimum (\frac{4}{e}, - \frac{56}{e})

f (x) = x^{2} - 6 x

f (x) = - 5 x^{2} + 2 x + 8

4 - 6 a

f (x) = \frac{- 2 x - 5}{6 x - 7}

f (x) = e^{- 2.5 x^{2}}

What is the extreme f(x)=7x^2ln(x/4) ?
The extreme f(x)=7x^2ln(x/4) is Minimum(4/(sqrt(e)),-(56)/e)