y = x*lnx

To determine the extreme points, first we need to determine first derivative;'s zeros.

y= x*lnx = u*v

u = x ==> u'= 1

v= ln x ==> v' = (1/x)

y'= u'v + uv';

= 1*lnx + x*(1/x)

= ln x + 1

ln x +1 = 0

==> ln x = -1

==> x= 1/e

Then, the function y has a crotical value at x= 1/e = e^-1

To find the y coordinates, let us substitute:

y= xlnx = (e^-1)ln e^-1= -e^-1 = -1/e

Then the extreme point is (1/e , -1/e)

To determine the coordinates of the extreme point of the given function, we'll differentiate the function.

f(x) = y = xlnx

Since the expression of the function is a product, we'll apply the product rule:

(u*v)' = u'*v + u*v'

We'll note u = x => u' = 1

We'll note v = ln x => v' = 1/x

u'*v + u*v' = 1*lnx + x*(1/x)

We'll eliminate like terms:

u'*v + u*v' = ln x + 1

To calculate the coordinates of the extreme point, we'll have to calculate the values of x for the derivative of the function is cancelling;

ln x + 1 = 0

ln x = -1

x = e^-1

x = 1/e

We'll calculate f(x) for x = 1/e

f(1/e) = (1/e)*ln e^-1

f(1/e) = -1*(1/e)*ln e

f(1/e) = -1/e

**The coordinates of the extreme point are: (1/e , -1/e).**

y = xlnx

To determine the extreme points.

Solution.

The extreme points of f(x) are obtained by setting f'(x) = 0.

y = f(x) = xlnx

f'(x) = (xlnx)' = (x)'lnx+ x(lnx)'

f'(x) = lnx +x/x = lnx +1

f"(x) = (lnx)' +(1)'

f"(x) = 1/x +0

So setting f'(x) = 0 gives lnx1 +1 = 0.

So lnx1 = -1 is a critical point.

x1 = e^(-1).

f''(x1) = 1/e^(-1) = e which is positive. This satisfies the condition for y to be minimum.

So at x = e^(-1) , f(x) xlnx has the minimum value y1 =e^(-1)*ln(e^-1) = - e^(-1).

So the required coordintes at this extreme value is (x,y) = (x1,y1) = {e^-1 , -e^(-1) }