Question

Verify Lagrange's Mean Value theorem for the given function: $f\left(x\right)=2x^2-7x+10$ in $[2,5]$.

Answer 1

Step 1. Definition of Lagrange's Mean Value theorem.

If a function $f$ is a continuous function on $[a,b]$ and differentiable on $(a,b)$. Then there exists a point $c$ in this interval $(a,b)$ such that the derivative of the function at the point $c$ is equal to the difference of the function values at these points, divided by the difference of the point values.

$f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Step 2. Verify Lagrange's Mean Value theorem for the given function.

Since a polynomial function is always continuous and differentiable into the given interval.

So, the function $f\left(x\right)=2x^2-7x+10$ is a continuous function on $[2,5]$ and differentiable on $(2,5)$.

$$\begin{gathered} \because f\left(x\right)=2x^2-7x+10 \\ \therefore \frac{d}{dx}f\left(x\right)=4x-7 \\ \Rightarrow f\text{'}\left(x\right)=4x-7 \\ \Rightarrow f\text{'}\left(c\right)=4c-7 \end{gathered}$$

From the definition of Lagrange's Mean Value theorem,

$$\begin{gathered} \therefore f\text{'}\left(c\right)=\frac{f\left(5\right)-f\left(2\right)}{5-2}\mathbf{}\left[\because f\left(5\right)=2\left(25\right)-7\times 5+10=25,f\left(2\right)=2\left(4\right)-7\left(2\right)+10=4\right] \\ \Rightarrow 4c-7=\frac{25-4}{3} \\ \Rightarrow 4c-7=7 \\ \Rightarrow c=\frac{14}{4} \\ \Rightarrow c=3.5\in (2,5) \end{gathered}$$

Hence, Lagrange's Mean Value theorem is verified for the given function.