The value of c in the Lagrange's Mean Value Theorem for the function f(x)=x^3-4x^2+8x+11, …

Question

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:

Accepted Answer

$f(0)=11,\ f(1)=16$. $f'(x)=3x^2-8x+8$.

$$f'(c)=\frac{16-11}{1}=5\Rightarrow3c^2-8c+8=5\Rightarrow3c^2-8c+3=0\Rightarrow c=\frac{8\pm2\sqrt7}{6}=\frac{4\pm\sqrt7}{3}$$

Since $c\in(0,1)$, $c=\dfrac{4-\sqrt7}{3}$.

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