# Finding the average value of a cubic function

Let $p(x)=ax^{3}+bx^{2}+cx+d$. There exists real numbers $r$ and $s$(independent of $a,b,c$ and $d$) $0<r<s<1$. For which the average value of $p(x)$ on the interval $[0,1]$ is equal to the average value of $p(r)$ and $p(s)$. Find the product of $rs$ expressed as a fraction.

What I know is $$f_{\text{avg}}=\frac{1}{b-a}\int\limits_a^bf(x)dx$$

$$f_{\text{avg}}=\int\limits_0^1f(x)dx$$

From that I am having trouble finding $r$ or $s$. I would think they would be the same but they clearly can't be. I know how to plug in the right values and take the integral but after that unsure. I'm having trouble understanding the question or what to try. Thanks for any help.

• Your expression for the average value is wrong. – preferred_anon Aug 9 '15 at 18:59
• I took the formula from wikipedia for the mean of a function over an interval. Or did I read the question wrong? – HighSchool15 Aug 9 '15 at 19:04
• @HighSchool15: in the average formula, you've replaced the limits of integration with the appropriate 0 and 1 for this problem, but you did not do so for the denominator, which in this case also has $b = 1$ and $a = 0$. This is particularly confusing since $a$ and $b$ have a different meaning in the problem (where they are coefficients of the polynomial, not limits of integration). – Paul Sinclair Aug 9 '15 at 19:09
• @HighSchool15 To be more specific, what are $a$ and $b$ in your formula for $f_{avg}$? – preferred_anon Aug 9 '15 at 19:10
• Try to compute the average (the integral). Try writing down $p(r)$ and $p(s)$ in terms of $a,b,c,d$. – preferred_anon Aug 9 '15 at 19:15

The question probably misses a condition $a,b,c,d\ne0$. We have \begin{align} p_{\mathrm{avg}} &= \tfrac14 a+\tfrac13 b+\tfrac12 c+d =p(r)+p(s) = \tfrac12 (r^3+s^3)a + \tfrac12 (r^2+s^2)b + \tfrac12(r+s)c +d \end{align}
Which gives $rs=\tfrac16$.
Moreover, we can find that $r=\tfrac12-\tfrac16\sqrt{3}$, $s=\tfrac12+\tfrac16\sqrt{3}$.
• Oh wow I miss read average of $p(r)+p(s)$ as the average value of each separately. Thank you very clear now. – HighSchool15 Aug 9 '15 at 19:30