Question

The value of c in the Lagrange’s mean value theorem for the function $f(x)={{x}^{3}}-4{{x}^{2}}+8x+11$ , where $x\in [0,1]$ is:
(a) $\dfrac{\sqrt{7}-2}{3}$
(b) $\dfrac{4-\sqrt{7}}{3}$
(c) $\dfrac{2}{3}$
(d) $\dfrac{4-\sqrt{5}}{3}$

Answer 1

Hint: Use the conditions of Lagrange’s Theorem according to which if a function f is continuous in the interval [a,b] and is differentiable in the interval (a,b) then there exists at least one c lying in the interval (a,b) such that $f'\left( c \right)=\dfrac{f(b)-f(a)}{b-a}$ . Also, the function given to us is a polynomial and polynomials are continuous and differentiable for all real values of x, so just find the differential of the function and put x=c in it and equate it with the value you get using $f'\left( c \right)=\dfrac{f(b)-f(a)}{b-a}$ .

Complete step-by-step answer:
Before starting with the solution, let us discuss Lagrange’s Theorem. The theorem states that if a function f is continuous in the interval [a,b] and is differentiable in the interval (a,b) then there exists at least one c lying in the interval (a,b) such that $f'\left( c \right)=\dfrac{f(b)-f(a)}{b-a}$ .
Now starting with the solution. The function given to us is $f(x)={{x}^{3}}-4{{x}^{2}}+8x+11$ in the interval [0,1] and it is a polynomial and we know that polynomials are continuous and differentiable for all real values of x. Therefore, we can say that:
$f'\left( c \right)=\dfrac{f(1)-f(0)}{1-0}............(i)$
Now we know $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$ , so, we can say:
$f'(x)=3{{x}^{2}}-8x+8$
Now we will put x=c. On doing so, we get
$f'(c)=3{{c}^{2}}-8c+8$
Now, if we put this in equation (i), we get
$3{{c}^{2}}-8c+8=\dfrac{f(1)-f(0)}{1-0}$
Now we will use the definition of the function f. On doing so, we get
$3{{c}^{2}}-8c+8=\dfrac{{{1}^{3}}-4\times {{1}^{2}}+8\times 1+11-\left( {{0}^{3}}-4\times {{0}^{2}}+8\times 0+11 \right)}{1-0}$
$\Rightarrow 3{{c}^{2}}-8c+8=\dfrac{1-4+8+11-11}{1}$
$\Rightarrow 3{{c}^{2}}-8c+8=5$
$\Rightarrow 3{{c}^{2}}-8c+3=0$
Now, the final equation we got was a quadratic equation. So, we will use the quadratic formula to get its root.
$\therefore c=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-(-8)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\times 3\times 3}}{2\times 3}=\dfrac{8\pm \sqrt{64-36}}{6}=\dfrac{8\pm \sqrt{28}}{6}$
But if we see $\dfrac{8+\sqrt{28}}{6}$ doesn’t lie in the rage (0,1), so the only possible value of c is $c=\dfrac{8-\sqrt{28}}{6}=\dfrac{2\left( 4-\sqrt{7} \right)}{6}=\dfrac{4-\sqrt{7}}{3}$
Therefore, the answer to the above question is option (b).

Note: While using Rolle’s Theorem and Lagrange’s mean value theorem, don’t forget to ensure that the function is differentiable and continuous in the given interval, as it is a necessary condition for this theorem to hold true. Also, don’t forget to make sure that the value of c is lying in the interval that you are using for these theorems.