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I am stumbling on an (at first sight) simple homework example. Maybe someone can help me. I just would need tips how to build up the equations for the respective polynomials, not the full solution.

Here is the homework example: An even polynomial of fourth degree $f(x)$ has a zero point at $(-2,0)$ and on another zero point it has a tangent with the equation $y = 6 - 6 \cdot x$. The polynomial of second degree $g(x)$ is intersecting with $f(x)$ at a maximum of $f(x)$, furthermore $g(x)$ itself has a maximum at $(0.5,5.25)$.

The task is now to determine the polynomials $f(x)$ and $g(x)$, so to determine their coefficients. So I have to determine the three coefficients of $f(x) = a_1 \cdot x^4 + a_2 \cdot x^2 + a_3$ and the three coefficients of $g(x) = b_1 \cdot x^2 + b_2 \cdot x + b_3$.

So how to get the needed three equations for determining $a_1,a_2,a_3$ and the three equations for determing $b_1,b_2,b_3$ ?

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Do it one step at a time.

$f(-2)=0$ gives you one equation.

If $f$ has $y=6-6x$ as a tangent at one of its zero crossings, then the zero crossings in question must be at the point where $y=6-6x$ crosses the $x$ axis. Setting $f(x)=0$ there gives you another equation. Finally, the slope of $f$ at that point must be $-6$, so setting $f'(x)=-6$ at that point gives you a third equation. Solve. Now you know $f$.

Alternatively, you know one zero of $f$; and another zero where the $6-6x=0$. Since $f$ is even, you can reflect those two zeroes in the $y$-axis to get two more, and then you have all the four possible zeroes of a fourth-degree polynomial. So $f(x)=c(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ for some $c$ -- but it turns out that setting $c=1$ already gives you the right slope at the point of tangency. How nice!

Now that you know $f$ you can sketch it and find that it has exactly one local maximum, which must be where it crosses $g$. That gives you one equation for the coefficients of $g$; knowing that $g(0.5)=5.25$ and $g'(0.5)=0$ gives you two more. Solve again!

Alternatively since you know the apex of $g$, you can be sure that $g(x)=d(x-0.5)^2+5.25$ for some factor $d$; then you just need to find the right $d$ such that $g$ intesects $f$ at the maximum, and multiply out.


(The alternatives at each step may seem more involved, but at least for me they're actually the simpler way. This way around I can keep everything in my head, up to and including finding $g(x)=-5x^2+5x+4$ -- whereas I wouldn't be able to do the three-equations-in-three-unknowns route without taking it to pencil and paper).

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First condition (given): $f(-2) = 0$.

Second condition (from evenness, may not be useful!): $f(2) = 0$.

Third condition: $f'(x_0) = 6$.

Fourth condition: $f(x_0) = 6-6x_0$.

Fifth condition: At some $x_1$, $f'(x_1) = 0$ and $g(x_1) = f(x_1)$.

Sixth condition: $g'(0.5) = 0, g(0.5) = 5.25$

Seventh condition (since $g$ has a maximum it must open downwards): $b_1 < 0$.

Can you use these to find what you need?

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  • $\begingroup$ @HenningMakholm That's what I intended. I meant that $x_1$ is a single specific $x_1$. But that may not be clear, let me edit! Thanks :) $\endgroup$ – Emily Jan 9 '15 at 19:03

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