The value of c in the Lagrange's mean value theorem for the function

Question:

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

 

  1. (1) $\frac{4-\sqrt{5}}{3}$

  2. (2) $\frac{4-\sqrt{7}}{3}$

  3. (3) $\frac{2}{3}$

  4. (4) $\frac{\sqrt{7}-2}{3}$


Correct Option: 2,

Solution:

Since, $f(x)$ is a polynomial function.

$\therefore \quad$ It is continuous and differentiable in $[0,1]$

Here, $f(0)=11, f(1)=1-4+8+11=16$

$f^{\prime}(x)=3 x^{2}-8 x+8$

$\therefore \quad f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}=\frac{16-11}{1}$

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

$\Rightarrow \quad 3 c^{2}-8 c+3=0$

$\Rightarrow \quad c=\frac{8 \pm 2 \sqrt{7}}{6}=\frac{4 \pm \sqrt{7}}{3}$

$\therefore \quad c=\frac{4-\sqrt{7}}{3} \in(0,1)$

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