Question

Let $f:[2,7]\to [0,\infty )$ be a continuous and differentiable function such that $\left(f\right(7)-f(2\left)\right)\frac{\left(f\right(7){)}^2+(f\left(2\right){)}^2+f\left(7\right)f\left(2\right)}{3}=k{f}^2\left(c\right)f^{\prime }\left(c\right),$ where $c\in (2,7).$ Then the value of $k$ is

Answer 1

We have $\left(f\right(7)-f(2\left)\right)\frac{\left(f\right(7){)}^2+(f\left(2\right){)}^2+f\left(7\right)f\left(2\right)}{3}=k{f}^2\left(c\right)f^{\prime }\left(c\right)$
$\Rightarrow f\left(7\right){)}^3-\left(f\right(2){)}^3=3k{f}^2\left(c\right)f^{\prime }\left(c\right)$
Let us consider the function $g\left(x\right)=\left(f\right(x){)}^3$ which is continuous in $[2,7]$ and differentiable in $(2,7)$.

Then from Lagrange's mean value theorem, there exists at least one $c\in (2,7)$ such that
$g^{\prime }\left(c\right)=\frac{g\left(7\right)-g\left(2\right)}{7-2}$
$\Rightarrow g^{\prime }\left(c\right)=\frac{\left(f\right(7){)}^3-(f\left(2\right){)}^3}{7-2}=3f^2\left(c\right)f^{\prime }\left(c\right)$
$\Rightarrow \left(f\right(7)-f(2\left)\right)\frac{\left(f\right(7){)}^2+(f\left(2\right){)}^2+f\left(7\right)f\left(2\right)}{3}=5f^2\left(c\right)f^{\prime }\left(c\right)$