Question

determine max/min values of function
f (x) = xlogx

Answer 1

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Hope this answer will help u..

Answer 2

Answer:

Minimum value is $f(e^{-1})=-\frac{1}{e}$

Step-by-step explanation:

Given : Function $f(x)=x\log x$

To find : Determine the max/min values of function ?

Solution :

To determine the max/min value double derivate the function.

$f(x)=x\log x$

Derivate w.r.t x,

$f'(x)=x\frac{d(\log x)}{dy}+\log x\frac{d(x)}{dy}$

$f'(x)=x\times \frac{1}{x}+\log x\times 1$

$f'(x)=1+\log x$

For max/min put derivate=0,

$1+\log x=0$

$\log x=-1$

$x=e^{-1}$

Now derivate w.r.t x again,

$f''(x)=0+\frac{1}{x}$

$f''(e^{-1})=\frac{1}{e^{-1}}$

$f''(e^{-1})=e>0$

i.e. f(x) is min at x.

The minimum value at $x=e^{-1}$ is

$f(e^{-1})=e^{-1}\log (e^{-1})$

$f(e^{-1})=\frac{1}{e}\times -1$

$f(e^{-1})=-\frac{1}{e}$