What are the critical values, if any, of $f(x)= (x-2)/(x^2-4)+2x$ ?

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Answer 1 · 2017-01-07T21:29:10

The critical points are:

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

We can simplify the function and remove one discontinuity noting that:

$(x^2-4) = (x-2)(x+2)$

hence:

$f(x) = (x-2)/(x^2-4)+2x = (x-2)/((x-2)(x+2))+2x =1/(x+2) +2x$

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

So we can find the critical points as roots of the equation:

$-1/((x+2)^2)+2 = 0$

$1/((x+2)^2)= 2$

$(x+2)^2 = 1/2$

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

As:

$f''(x) = 2/((x+2)^3)$

we have:

$f''(-2- 1/sqrt(2)) = 2/((-2- 1/sqrt(2)+2)^3)=-4sqrt(2) <0$

$f''(-2+ 1/sqrt(2)) = 2/((-2+ 1/sqrt(2)+2)^3)=4sqrt(2) >0$

so $x=-2- 1/sqrt(2)$ is a local maximum and $x=-2+ 1/sqrt(2)$ is a local minimum.

graph{(x-2)/(x^2-4)+2x [-23.13, 16.87, -13.6, 6.4]}

What are the critical values, if any, of f(x)= (x-2)/(x^2-4)+2x f(x)= (x-2)/(x^2-4)+2x?