1. Tardigrade
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  3. Mathematics
  4. Let f:[2,7] arrow[0, ∞) be a continuous and differentiable function. Then, (f(7)-f(2)) ((f(7))2+(f(2))2+f(2) f(7)/3) is, where c ∈[2,7] .

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Q. Let $f:[2,7] \rightarrow[0, \infty)$ be a continuous and differentiable function. Then, $(f(7)-f(2)) \frac{(f(7))^{2}+(f(2))^{2}+f(2) f(7)}{3}$ is, where $c \in[2,7] .$

Continuity and Differentiability

Solution:

Let $g(x)=f^{3}(x) $
$\Rightarrow g'(x)=3 f^{2}(x) \cdot f'(x)$
$\because f:[2,7] \rightarrow[0, \infty)$
$ \Rightarrow g:[2,7] \rightarrow[0, \infty)$
Using Lagrange's mean value theorem on $g(x)$, we get
$g'(c)=\frac{g(7)-g(2)}{5}, c \in[2,7]$
$\Rightarrow 5 f^{2}(c) f'(c)=(f(7)-f(2)) \frac{(f(7))^{2}+(f(2))^{2}+f(2) f(7)}{3}$