Mathematics - Quadratic Equation Question with Solution | TestHub

The function $\varphi (x)$ is constructed such that $\varphi (a)=\varphi (b)=\varphi (c)=0$. By Rolle's Theorem, there exist $\alpha \in (a,c)$ and $\beta \in (c,b)$ such that ${\varphi }^{\prime }(\alpha )=0$ and ${\varphi }^{\prime }(\beta )=0$. Applying Rolle's Theorem again to ${\varphi }^{\prime }(x)$ on $(\alpha ,\beta )$, there exists $\mu \in (\alpha ,\beta )$ (and thus $\mu \in (a,b)$) such that ${\varphi }^{\prime \prime }(\mu )=0$.

Let's compute ${\varphi }^{\prime \prime }(x)$.

So, $\varphi (x)=P(x)-f(x)$.

$$P^{\prime \prime }(x)=\frac{f(a)}{(a-b)(a-c)}\cdot 2+\frac{f(b)}{(b-c)(b-a)}\cdot 2+\frac{f(c)}{(c-a)(c-b)}\cdot 2$$

This is because the terms like $\frac{(x-b)(x-c)}{(a-b)(a-c)}$ are quadratic in $x$, and their second derivative is $2\cdot \frac{1}{(a-b)(a-c)}$.

Since ${\varphi }^{\prime \prime }(\mu )=0$

$$\frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-c)(b-a)}+\frac{f(c)}{(c-a)(c-b)}=\frac{1}{2}{f}^{\prime \prime }(\mu )$$