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[SOLVED] find inflection pt with constant K

karush

Well-known member
Jan 31, 2012
2,722
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
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
$f''(x)=0$ 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
Jan 31, 2012
2,722
$f''(x)=0$ 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