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  • The value of c in the Lagrange's Mean Value Theorem for the function <img alt="\mathrm{f(x)=x^3-4 x^2+8 x+13 }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3Dx%5E3

The value of c in the Lagrange's Mean Value Theorem for the function \mathrm{f(x)=x^3-4 x^2+8 x+13 }, when \mathrm{x \in[0,1] } is

Option: 1

\frac{2}{3}


Option: 2

\frac{4-\sqrt{7}}{3}


Option: 3

\frac{4+\sqrt{7}}{3}


Option: 4

\frac{\sqrt{7}-2}{3}


Answers (1)

best_answer

Since f(x) is a polynomial function.
\mathrm{\therefore \quad } It is continuous and differentiable in [0,1]
Here, f(0)=13, f(1)=1-4+8+13=18
\mathrm{f^{\prime}(x)=3 x^2-8 x+8 }
\mathrm{\begin{aligned} & \therefore f^{\prime}(c)=\frac{f(1)-f(0)}{1-0} \Rightarrow 3 c^2-8 c+8=5 \\ & \Rightarrow 3 c^2-8 c+3=0 \Rightarrow 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) \end{aligned} }

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