Solve Quartic Polynomial: Find Coefficients & Values of q - Yahoo! Answers

[MarkFL]

Here is the question:

Let P(x) = x4 + ax3 + bx2 + cx + d. Need to find the values of the variable that satisfy the guidelines below?

Let P(x) = x4 + ax3 + bx2 + cx + d. The graph of y = P(x) is symmetric with respect to the y-axis, has a relative MINIMUM at (0, 1) and has an absolute minimum at (q, -3). Determine the values of a,b,c and d, using these values write an expression for p(x). Also, find all possible values of q.

Here is a link to the question:

Let P(x) = x4 + ax3 + bx2 + cx + d. Need to find the values of the variable that satisfy the guidelines below? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello tin h,

We are given:

\(\displaystyle P(x)=x^4+ax^3+bx^2+c^x+d\)

If $P(x)$ is symmetric with respect to the $y$-axis, then this means $P(x)$ is even, i.e., $P(-x)=P(x)$, and so this means $a=c=0$, leaving us with:

\(\displaystyle P(x)=x^4+bx^2+d\)

The fact that $P(x)$ has a relative minimum at $(0,1)$ tells us:

\(\displaystyle P(0)=d=1\)

\(\displaystyle P'(x)=4x^3+2bx\)

The fact that \(\displaystyle P'(0)=0\) yields no additional information.

\(\displaystyle P''(x)=12x^2+2b\)

We know by the second derivative test we must have:

\(\displaystyle P''(0)>0\,\therefore\,0<b\).

The fact that $P(x)$ has an absolute minimum at $(q,-3)$ tells us:

\(\displaystyle P(q)=q^4+bq^2+1=-3\,\therefore\,q^4+bq^2=-4\)

\(\displaystyle P'(q)=4q^3+2bq=0\,\therefore\,2q^3+bq=0\)

Solving both equations for $b$ and equating, we find:

\(\displaystyle b=-\frac{q^4+4}{q^2}=-2q^2\)

\(\displaystyle q^4+4=2q^4\)

\(\displaystyle q^4=4\)

and since $q$ is real, we find:

\(\displaystyle q=\pm\sqrt{2}\)

and so substituting into the first equation we used, we find:

\(\displaystyle b=-4\)

\(\displaystyle P''(q)=12q^2+2b\)

We know by the second derivative test we must have:

\(\displaystyle P''(0)>0\,\therefore\,0<6q^2+b\).

We see that the values we found for $b$ and $q$ satisfy this requirement.

However, we find that the extremum at (0,1) must be intended to be a local maximum, not a minimum.

In conclusion, we have found:

\(\displaystyle a=0,\,b=-4,\,c=0,\,d=1,\,q=\pm\sqrt{2}\)

To tin h and any other guests viewing this topic, I invite and encourage you to register and post other calculus problems in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.