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Problem of the Week #97 - February 3rd, 2014

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: For what values of $c$ is there a straight line that intersects the curve \[y=x^4+cx^3+12x^2-5x+2\]
in four distinct points?

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's problem was correctly answered by anemone, magneto, MarkFL, and mente oscura. You can find anemone's solution below.

If there is a straight line that will intersect the curve $y=x^4+cx^3+12x^2-5x+2$ at four distinct points, that means the curve of $y$ must have three critical points or more precisely, two inflexion points.

From the definition of inflexion point, we know we can find its two x-coordinates of the inflexion point by setting the second derivative of the original function as zero. In this instance, the second derivative of the quartic function yields a quadratic function and if we solve for $x$ of the quadratic function by letting it equals zero, we must obtain two distinct $x$ values. This is achievable iff the discriminant is greater than zero.

$y'=4x^3+3cx^2+24x-5$

$y''=12x^2+6cx+24$

The discriminant of the second derivative must be greater than zero, i.e. $(6c)^2-4(12)(24)>0$. Solving this inequality for $c$, we will find the range of value of $c$ that will allow a straight line to cut the curve $y=x^4+cx^3+12x^2-5x+2$ at four distinct points.

$(6c)^2-4(12)(24)>0$

$2^23^2c^2-2^73^2>0$

$c^2-2^5>0$

$c^2-4^2(2)>0$

$(c-4\sqrt{2})(c+4\sqrt{2})>0$

This gives the solution set of $c$ as:

${c: c<-4\sqrt{2}\; \text{or}\; c>4\sqrt{2}}$
 
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