# f(x)=2ln(x^2+3)-x with domain -3<=x<=5find the x-coordinates of each relative maximum point and each relative minimum point of f and find the x coordinates of each inflection point of f

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`f(x) = 2ln(x^2+3)-x` ` -3lt=xlt=5`

For extreme points (maxima, minima and inflection points) f'(x) = 0.

`f'(x) = 2((2x)/(x^2+3)) -1`

`f'(x) = (4x)/(x^2+3) -1`

For extreme points, `f'(x) = 0` , then,

`(4x)/(x^2+3) -1 = 0`

`4x -(x^2+3) = 0`

This gives us a quadratic equation,

`x^2-4x+3 = 0`

`(x-1)(x-3) = 0`

`x =1` or `x =3`

We have two points at which we have extreme conditions. To determine whether these are maxima, minima and inflection points, we have to check the sign of the second derivative, f''(x).

`f''(x) = (4(x^2+3) -4x(2x))/(x^2+3)^2`

`f''(x) = (4(x^2+3-2x^2))/(x^2+3)^2`

`f''(x) = (4(3-x^2))/(x^2+3)^2`

At x =1,` f''(x) gt 0` . Therefore f(x) has a relative minimum point at x =1.

Then, at `x=3` , `f''(x) lto` . Therefore f(x) has a relative maximum point at `x =3`

There are no inflection points.

**Relative maximum at x =3**

**Relative minimum at x= 1**