MHB Find inflection pt with constant K

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karush
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for c I know inflection pts are found from $$f''(x)$$ but since I didn't know at what value $$x$$ would be I didn't know how to find $$k$$
also I assume on the $$x$$ 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
 
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$f''(x)=0$ 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 $k$.
 
Ackbach said:
$f''(x)=0$ 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 $$k$$ change the shape of the curve, there seems to a IP at $$x=16$$ but that is when k=1 and it doesn't go thru the $$x$$ axis

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

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