# find all the critical points of the function f(x)= (e^x)/(x^2-1)

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### 1 Answer

`f(x) = (e^x)/(x^2-1)`

To find the critical points we have to find the derivative of f(x)

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

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

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

At critical points,

`f'(x) = 0`

This means,

`x^2-2x-1 = 0` (we know `e^x != 0` )

This gives,

`x = (+2+-sqrt(4+4))/(2xx1)`

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

`x = 1+-sqrt(2)`

Therefore, the x coordinates of critical points are,

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

At `x = 1-sqrt(2)` ,

`f(x) = e^(1-sqrt(2))/((1-sqrt(2))^2-1)`

`f(x) = e^(1-sqrt(2))/(1-2sqrt(2)+2-1)`

`f(x) = e^(1-sqrt(2))/(2-2sqrt(2))`

At `x = 1+sqrt(2),`

`f(x) = e^(1+sqrt(2))/((1+sqrt(2))^2-1)`

`f(x) = e^(1+sqrt(2))/(1+2sqrt(2)+2-1)`

`f(x) = e^(1+sqrt(2))/(2+2sqrt(2))`

Therefore the critical points are,

`(1-sqrt(2), e^(1-sqrt(2))/(2-2sqrt(2)))`

and

`(1+sqrt(2), e^(1+sqrt(2))/(2+2sqrt(2)))`