# [SOLVED]find inflection pt with constant K

#### karush

##### Well-known member
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

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#### Ackbach

##### Indicium Physicus
Staff member
$f''(x)=0$ 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

##### Well-known member
$f''(x)=0$ 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