# How many extreme points does the function y=3lnx - 3x^2 have?

mathsworkmusic | Certified Educator

We are given the function

`y = 3lnx - 3x^2`

and asked to fund the number of extreme points. An extreme point is an absolute maximum or absolute minimum, absolute meaning that it is finite in value.

To find any extreme points we set the derivative of the function to zero. The nature of extreme points is that the gradient is zero at that point, the gradient being also the first derivative of the function. So, we have that the derivative with respect to `x` is given by

`(dy)/(dx) = y' = 3/x - 6x`

(the derivative of `lnx` is `1/x` and of `x^a = ax^(a-1)` )

Setting this to zero we will find ` `` `values for `x` corresponding to extreme points:

`3/x - 6x = 0`  ` `implies  `6x = 3/x`  implies `x^2 = 3/6 = 1/2` .

Therefore, values for `x` corresponding to extreme values will satisfy `x^2 = 1/2` . This would be `x = pm 1/sqrt(2) = pm 0.71`.

To find the corresponding `y` values, plug these values into the original function so that `y =`

`3ln(1/sqrt(2)) - 3(1/sqrt(2))^2 = -1.040 - 3/2 = -2.54`

or `3ln(-1/sqrt(2)) - 3(-1/sqrt(2))^2 = ` undefined! Because log of a negative number is undefined.

Therefore there is only one extreme point on the function `y = 3ln(x) - 3x^2` , at `(x,y) = (0.71,-2.54)`. The other possibility for the value of `x` was out of the domain of the function, which is `(0,infty)`.` ` This can be seen when looking at the graph of the function

Also looking at this we can see that the extreme point is a maximum. There are not absolute minima as the graph goes to minus infinity at both ends (a non-finite value).

We could also show that the extreme point is a minimum by checking the second derivative of the function

`y'' = -3/x^2 - 6`

Since this is negative at our extreme point (`x= 0.71`), the gradient is decreasing at that point, meaning that it goes from positive /, through zero -, to negative \ .

Answer: there is only 1 extreme point (a maximum), which is at (x,y) = (0.71,-2.54). The other candidate x value for an extreme point, -0.71, is out of the domain of the function and so does not qualify.

aruv | Student

`y=3ln(x)-3x^2`

`y'=3/x-6x`

`` For extrema we have

`3/x-6x=0`

`x^2=1/2`

`x=+-1/sqrt(2)`

But

`x=-1/sqrt(2)`  does not lies in domain of definition of function y(x).

Thus function y(x) has only one extrem point which is global maxima.

`x=1/sqrt(2)`

`y=3ln(2^(-1/2))-3(1/2)`

`=-3/2ln(2)-3/2=-3/2(1+ln(2))`

Thus

point

`(1/sqrt(2),-(3/2)(1+ln(2)))` is the only point of extrema which is appear in  graph below where graph is turning .