Question

Use Rolle's theorem to prove that equation $a{x}^2+bx=\frac{a}{3}+\frac{b}{2}$ has a root between 0 and 1.

Answer 1

we know that According to the Rolle's Theorem

$F\left(x\right)$ should continuous in $\left[a,b\right]$

$F\left(x\right)$ should differentiable in $(a,b)$

And $F\left(a\right)=F\left(b\right)$

Then, there exist at least one value of $x$, $\left(letC\right)$ where $C\in \left(a,b\right)$ such that $F^{\prime }\left(C\right)=0$

Let

$\begin{array}{c}{F}^{\prime }\left(x\right)=a^{x^2}+bx-\left(\frac{a}{3}+\frac{b}{2}\right)\\ F\left(x\right)=\int \left[a{x}^2+bx-\left(\frac{a}{3}+\frac{b}{2}\right)\right]dx\\ =\frac{a{x}^3}{3}+\frac{b{x}^2}{2}-\left(\frac{a}{3}+\frac{b}{2}\right)x+C\end{array}$

Now,

$F\left(x\right)$ is continuous in $\left[0,1\right]$ as it is polynomial

$F\left(x\right)$ is also differentiable in $(0,1)$

$\begin{array}{c}F\left(0\right)=C\\ F\left(1\right)=\frac{a}{3}+\frac{b}{2}-\left(\frac{a}{3}+\frac{b}{2}\right)+C=C\\ F\left(0\right)=F\left(1\right)\end{array}$

Hence,

All conditions are fulfilled, so there exist a ${}^{\prime }{C}^{\prime }$ so that $F^{\prime }\left(x\right)=a{x}^2+bx-\left(\frac{a}{3}+\frac{b}{2}\right)$ has a root.