Find inflection pt with constant K

[karush]

View attachment 2095
for c I know inflection pts are found from \(\displaystyle f''(x)\) but since I didn't know at what value \(\displaystyle x\) would be I didn't know how to find \(\displaystyle k\)
also I assume on the \(\displaystyle x\) axis means the graph either touches or crosses the graph at IP.

(image of typing is mine) not sure if this in the right forum

 

[Ackbach]

$f''(x)=0$ is a necessary, but not sufficient, condition for $x$ to be a point of inflection. The second derivative must change sign at a twice-differentiable point in order to guarantee a point of inflection. Try this and see if this imposes any conditions on $k$.

 

[karush]

$f''(x)=0$ is a necessary, but not sufficient, condition for $x$ to be a point of inflection. The second derivative must change sign at a twice-differentiable point in order to guarantee a point of inflection. Try this and see if this imposes any conditions on $k$.

not sure if I understand what a "twice-differentialbe point is"

also, doesn't \(\displaystyle k\) change the shape of the curve, there seems to a IP at \(\displaystyle x=16\) but that is when k=1 and it doesn't go thru the \(\displaystyle x\) axis

I can't seem to get both k and x to work for the IP