I am given this graph and I am asked to find the range of values of k such that the graph has no stationary points.
$$y=\frac{(x+2)}{x(x+k)}$$
I understand to differentiate the equation and let it “equate” to 0 (reason I put use “” is cause technically the equation cannot equate to 0 because it doesn’t have a solution) to find an equation to apply discriminatory<0 to find the range of values of k.
I would eventually get $k>2$. But the next step is to let $k=2$ and sub back into y to find that I would get a y=1/x graph with no stationary point. So $k\geq 2$. But I do not understand why my discrimination rule does not cover the full range of values for k? is this just a coincidence that subbing 2 into y will get me a graph with no stationary point or is there a reasoning behind it? and how else would I be able to spot such a thing next time?
Yes your derivation is right indeed, we have that for $x\neq -k$
$$y=\frac{(x+2)}{x(x+k)} \implies y'=\frac{x(x+k)-(2x+k)(x+2)}{x^2(x+k)^2}$$
and by $y'\neq 0$ we obtain
$$x(x+k)-(2x+k)(x+2)\neq 0$$
$$x^2+4x+2k\neq 0$$
and the latter requires that
$$\Delta=16-8k <0 \implies k>2$$
Now by inspection we see that the case $k=2$ also leads to a graph with no stationary point. This happens because for $k=2$ the original equation becomes
$$\frac{(x+2)}{x(x+2)}=\frac1x$$
and the cancellation of the $(x+2)$ term requires $x\neq -2$ and therefore the condition for the second order equation
$$x^2+4x+4=(x+2)^2\neq 0$$
always holds and this fact is consistent with the result obtained. In other words the discriminant would be zero but the quadratic equation has not zeros because the value $x=-2$ is not allowed.