The value of $$c$$, in the Lagrange's mean value theorem for the function $$f(x) = x^3 - 4x^2 + 8x + 11$$, when $$x \in [0, 1]$$, is

For applying Lagrange’s Mean Value Theorem we start with its statement: for a function which is continuous on a closed interval $$[a,b]$$ and differentiable on the open interval $$(a,b)$$, there exists at least one point $$c \in (a,b)$$ such that

$$f'(c)=\dfrac{f(b)-f(a)}{b-a}.$$

Here we have $$f(x)=x^{3}-4x^{2}+8x+11$$ defined on $$[0,1]$$, so we take $$a=0$$ and $$b=1$$. The given polynomial is continuous and differentiable everywhere, hence all the conditions are satisfied.

First we compute the values of the function at the endpoints:

$$\begin{aligned} f(1) &= 1^{3}-4(1)^{2}+8(1)+11 \\ &= 1-4+8+11 \\ &= 16, \\ f(0) &= 0^{3}-4(0)^{2}+8(0)+11 \\ &= 11. \end{aligned}$$

Now the average rate of change (the slope of the chord) on $$[0,1]$$ is

$$\dfrac{f(1)-f(0)}{1-0}=\dfrac{16-11}{1}=5.$$

Next we find the derivative of the function:

$$\begin{aligned} f(x)&=x^{3}-4x^{2}+8x+11,\\ f'(x)&=3x^{2}-8x+8. \end{aligned}$$

By the theorem, at the required point $$c$$ we must have

$$f'(c)=5.$$

So we set

$$3c^{2}-8c+8=5.$$

Bringing all terms to one side gives

$$3c^{2}-8c+8-5=0,$$

which simplifies to

$$3c^{2}-8c+3=0.$$

We solve this quadratic equation. The quadratic formula states that for $$Ax^{2}+Bx+C=0$$ the roots are

$$x=\dfrac{-B\pm\sqrt{B^{2}-4AC}}{2A}.$$

Here $$A=3,\;B=-8,\;C=3$$, so

$$\begin{aligned} c &=\dfrac{-(-8)\pm\sqrt{(-8)^{2}-4\cdot3\cdot3}}{2\cdot3}\\ &=\dfrac{8\pm\sqrt{64-36}}{6}\\ &=\dfrac{8\pm\sqrt{28}}{6}. \end{aligned}$$

Because $$\sqrt{28}=2\sqrt{7}$$, this becomes

$$c=\dfrac{8\pm2\sqrt{7}}{6}=\dfrac{4\pm\sqrt{7}}{3}.$$

Thus we obtain two possible values:

$$c_{1}=\dfrac{4+\sqrt{7}}{3},\qquad c_{2}=\dfrac{4-\sqrt{7}}{3}.$$

We now check which of these lies inside the open interval $$(0,1)$$:

$$\begin{aligned} c_{1}&=\dfrac{4+\sqrt{7}}{3}\approx\dfrac{4+2.6458}{3}\approx2.215\; >1,\\ c_{2}&=\dfrac{4-\sqrt{7}}{3}\approx\dfrac{4-2.6458}{3}\approx0.451\; \in(0,1). \end{aligned}$$

Only $$c=\dfrac{4-\sqrt{7}}{3}$$ satisfies $$0<c<1$$, so this is the required point ensured by Lagrange’s Mean Value Theorem.

Hence, the correct answer is Option B.