<s>[INST] Find a value b for the average value of $f(x)=-2x^2+3x+4$ over [0, b] Find a real number b, if it exists, for which the average value of $f(x)=-2x^2+3x+4$ over [0, b] is equal to 5. In other words, on the interval [0,b], there is an average value = 5 for the function $f(x)=-2x^2+3x+4$. I'm mainly checking my answer here. After using the average value theorem to find the integral (this is a rather trivial integral), I get the equation $5=\frac{1}{b}(\frac{-2}{3}x^3+\frac{3}{2}x^2+4x)|^{b}_{a}$. Now how do I show whether $b$ exists or not?
 [/INST] Since your interval is of the form $[0,b]$, this means that if the average value of $f(x)$ over said interval is $5$, then you get the equation
$$
5 = \frac{1}{b-0}\int_{0}^{b} -2x^2 +3x+4 \ dx
$$
Which, as you pointed out, gives you
\begin{align*}
5 = \frac{1}{b} \left(-\frac{2}{3}x^3 + \frac{3}{2} x^2 + 4x \right)\Bigg\vert_{0}^{b} = -\frac{2}{3}b^2 + \frac{3}{2} b + 4
\end{align*}
And rewriting the above equation we find that
$$
-\frac{2}{3}b^2 + \frac{3}{2} b -1 = 0 \implies 4\color{blue}{b^2}-9\color{blue}{b}+6 =0
$$
which is a quadratic equation in terms of $b$. We can use the quadratic formula to solve for $b$ by doing
$$
b = \frac{9 \pm \sqrt{(-9)^2 -4(4)(6)}}{2(4)}
$$
but since
$$
 \sqrt{(-9)^2 -4(4)(6)} =  \sqrt{-15}
$$
we find that the solutions for $b$ are complex since they involve the square root of a negative number! This means that there doesn't exist a real number $b$ such that the average value of $f(x) = -2x^2 +3x+4$ is $5$.
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