# Find all values of $c$ such that there exists a line that intersects the graph of $f(x) =x^4+9x^3+cx^2+9x+4$ in four distinct points.

Find all values of $c$ such that there exists a line that intersects the graph of $f(x) =x^4+9x^3+cx^2+9x+4$ in four distinct points.

I considered the derivative $f'(x)=4x^3+27x^2+2cx+9$. If it could be written in the form $4(x^2+r^2)(x-s)$ or $4(x-t)^3$, then there's only one local minimum. The first yields $r^2=1/3,s=-27/4,c=2/3$ and the second yields no solution. If $c=2/3$, then $f$ is strictly increasing for $x>-27/4$ and strictly decreasing for $x<-27/4$. Therefore if $c=2/3$ then there wouldn't exist such line.

However, the correct answer to this problem is $c<243/8.$ What's wrong with the claim $c\neq2/3$?

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What contest? Also, have you correctly transcribed the problem? – Will Jagy Nov 23 '12 at 21:14
Are you considering only horizontal lines that might intersect your graph? – Lubin Nov 23 '12 at 21:20
It's from the Putnam Competition. I didn't restrict my attention on horizontal lines. I'm not seeing where's wrong with my argument about c=2/3. – Christmas Bunny Nov 23 '12 at 21:28
What's wrong with $c=2/3$? Just because the graph has only one minimum, I see no reason why a nonhorizontal line couldn’t intersect it four times. – Lubin Nov 23 '12 at 21:48
Indeed, Wolfram’s picture for $c=2/3$ already shows four collinear points. – Lubin Nov 23 '12 at 22:02

The condition that the graph of $f$ should have no four points collinear is equivalent to the condition that the graph should have no real inflection points. Now, the second derivative is $f''=2\lbrack6x^2+27x+c\rbrack$, and we need the discriminant of what's in the brackets to be positive. The discriminant is $\Delta=729-24c$, and its positivity-condition is $c\lt243/8$.