Finding Stationary Points in a Rational Function

[HeyHow!]

i have been give the equation

f(x) = 1/(x+3) + 1/(x-1)

i am trying to find the stationary points. so

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

f '(x) = 0 for stationary point

then when i simplified down i get 2x^2+4x+5 = 0, which is undefinable

i looked at the derivative on the graph and it looks like it has a turning point at (-1,0).

please help, either i have made an easy mistake or somethings wrong!

 

[Muzza]

Yeah, looks like the derivative has a "turning point" around x = -1. However, that doesn't mean that the derivative IS zero at that point, it means that the derivative of the derivative is zero there. If you wanted to x such that f'(x) = 0, you'd look for the places where f'(x) crosses the x-axis (which, in this case, it doesn't appear to do).

 

[marlon]

Shouldn't it be 2x^2 + 4x + 10 =0. this has no real solution so the function is always increasing.

You got to look at the second derivative (f'') of this function and solve f'' = 0. Then you will get the x-coördinate of the turning point. Substituting this x-value into f(x) yields the y-value.


regards
marlon

 

[arildno]

marlon:
The function is always decreasing, not increasing!
The function doesn't have any stationary points.
The function has three continuous sections , each of them decreasing; the intervals where the function is defined are:
$$-\infty<{x}<{-3},-3<{x}<{1},1<x<\infty$$

 

[marlon]

marlon:
The function is always decreasing, not increasing!

yes that is correct. Forgot the minus sign in the first derivative...

Thanks for the correction.

regards
marlon