Question

Verify Lagrange’s Mean Value Theorem for the function $\mathrm f\left(\mathrm x\right)=\sqrt{\mathrm x^2-\mathrm x}$ in the interval [1,4].

Answer 1

Hint: First prove that the function is continuous and differentiable and then verify the theorem using the formula-

Complete step-by-step answer:

Substitute values of $a$ and $b$ given in question in $f(x)$ to find $f(a)$ and $f(b)$. To find the value of $f’(c)$, differentiate $f(x)$ with respect to $x$ and substitute $c$ to find $f’(c)$.

The Lagrange’s Mean Value Theorem(MVT) states that if a function $f(x)$ is continuous on a closed interval $[a,b]$ and differentiable on an open interval $(a,b)$, then there is at least one point $x=c$ on this interval such that,

$\mathrm f\left(\mathrm b\right)-\mathrm f\left(\mathrm a\right)=\mathrm f'\left(\mathrm c\right)\left(\mathrm b-\mathrm a\right)$ where $a<=c<=b$
For this problem, we have been given the value of $f(x)$ and the values of $a$ and $b$.
$a=1$ and $b=4$. To verify MVT, we have to prove that the function is continuous in closed interval $[1,4]$ and differentiable in open interval $(1,4)$.

The test for continuity is as follows-
A polynomial function is differentiable and continuous at all points. Since the given function is a square root of a polynomial function $x^2-x$ , $f(x)$ is continuous at every point between 1 and 4. Now we will check if $f(x)$ is defined at $x=1$ and $x=4$. If it is defined, the function is continuous.
$f\left(1\right)=\sqrt{1^2-1}=\sqrt0=0\;....\left(1\right)\\f\left(4\right)=\sqrt{4^2-4}\\=\;\sqrt{16-4}\\=\sqrt{12}=\sqrt{2\times2\times3}=2\sqrt3....\left(2\right)$
$f(x)$ is defined at all values of $x$ and is a polynomial function, so it is continuous.

The test for differentiability is as follows-
A polynomial function is differentiable and continuous at all points. Since the given function is a square root of a polynomial function $x^2-x$ , $f(x)$ is differentiable at every point between 1 and 4. Hence, $f(x)$ will be differentiable in the open interval $(1,4)$.

The final step is to verify the Lagrange’s Mean Value Theorem for $f(x)$. We know that,
$\mathrm f\left(\mathrm b\right)-\mathrm f\left(\mathrm a\right)=\mathrm f'\left(\mathrm c\right)\left(\mathrm b-\mathrm a\right)$where $b=4, a=1$
Substituting the values of $a$ and $b$,
$\mathrm f\left(4\right)-\mathrm f\left(1\right)=\mathrm f'\left(\mathrm c\right)\left(4-1\right)$
We have already calculated the values of $f(1)$ and $f(4)$ in equations (1) and (2).
So,
$2\sqrt3-0=\mathrm f'\left(\mathrm c\right)\left(3\right)\\\mathrm f'\left(\mathrm c\right)=\dfrac{2\sqrt3}3........\left(3\right)$
To find $f’(c)$, we will differentiate $f(x)$ with respect to $x$ and then substitute the value of $c$.
We can use the chain rule which is-
$\dfrac{\operatorname dy}{\operatorname dx}=\dfrac{\operatorname dy}{\operatorname dt}\times\dfrac{\operatorname dt}{\operatorname dx}$

Let $y = f(x)$, and differentiate $y$ with respect to $x$-
$\dfrac{\operatorname d\mathrm y}{\operatorname d\mathrm x}=\dfrac{\operatorname d\sqrt{\mathrm x^2-\mathrm x}}{\operatorname d\mathrm x}\\=\dfrac{\operatorname d\sqrt{\mathrm x^2-\mathrm x}}{\operatorname d\left(\mathrm x^2-\mathrm x\right)}\times\dfrac{\operatorname d\left(\mathrm x^2-\mathrm x\right)}{\operatorname d\mathrm x}\\=\dfrac1{2\sqrt{\mathrm x^2-\mathrm x}}\times\left(2\mathrm x-1\right)\\=\mathrm f'\left(\mathrm x\right)\\\\\mathrm f'\left(\mathrm c\right)=\dfrac1{2\sqrt{\mathrm c^2-\mathrm c}}\times\left(2\mathrm c-1\right).....\left(4\right)$

We will equate equations (3) and (4) to find the value of $c$.
$\dfrac{2\mathrm c-1}{2\sqrt{\mathrm c^2-\mathrm c}}=\dfrac{2\sqrt3}3\\3\left(2\mathrm c-1\right)=2\sqrt3\times2\sqrt{\mathrm c^2-\mathrm c}\\\mathrm{Squaring}\;\mathrm{both}\;\mathrm{sides},\\9\left(2\mathrm c-1\right)^2=4\times3\times4\left(\mathrm c^2-\mathrm c\right)\\9\left(4\mathrm c^2+1-4\mathrm c\right)=48\left(\mathrm c^2-\mathrm c\right)\\36\mathrm c^2+9-36\mathrm c=48\mathrm c^2-48\mathrm c\\12\mathrm c^2-12\mathrm c-9=0\\4\mathrm c^2-4\mathrm c-3=0\\4\left(\mathrm c^2-\mathrm c-\dfrac34\right)=0\\\mathrm c^2-\dfrac32\mathrm c+\dfrac12\mathrm c-\dfrac34=0\\\left(\mathrm c-\dfrac32\right)\left(\mathrm c+\dfrac12\right)=0\\c=\dfrac32,\;-\dfrac12\\$
We can reject $x=-\dfrac{1}{2}$ as it lies outside the interval. But
$1\leq\dfrac32\leq4$
Hence Langrange’s Mean Value Theorem is verified.

Note: For solving such types of problems one should have complete knowledge of MVT, one common mistake is that students verify continuity and differentiability of the function and jump straight to verification of LMVT. Polynomial functions are continuous and differentiable at all values. This fact can be used directly which will save a lot of time.