(D) Since $f$ is a polynomial function,it is continuous on $[2,7]$ and differentiable on $(2,7)$.According to the Lagrange's Mean Value Theorem,there

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(D) Since $f$ is a polynomial function,it is continuous on $[2,7]$ and differentiable on $(2,7)$.According to the Lagrange's Mean Value Theorem,there exists some $c \in (2,7)$ such that:$\frac{f(7)-f(2)}{7-2} = f^{\prime}(c)$Given that $f^{\prime}(x) \leq 5$ for all $x \in (2,7)$,we have $f^{\prime}(c) \leq 5$.Substituting the values,we get:$\frac{f(7)-3}{5} \leq 5$$f(7)-3 \leq 25$$f(7) \leq 28$Thus,the maximum possible value of $f(7)$ is $28$.

Class 12 Mathematics Continuity and Differentiation Rolle’s theorem, Lagrange's mean value theorem

Medium

Let $f$ be a polynomial function defined on $[2,7]$. If $f(2)=3$ and $f^{\prime}(x) \leq 5$ for all $x$ in $(2,7)$,then the maximum possible value attained by $f$ at $x=7$ is

AP EAMCET 2019 All AP EAMCET

A
$7$
B
$14$
C
$18$
D
$28$

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Continuity and Differentiation

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Rolle’s theorem, Lagrange's mean value theorem

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Let $f$ be a polynomial function defined on $[2,7]$. If $f(2)=3$ and $f^{\prime}(x) \leq 5$ for all $x$ in $(2,7)$,then the maximum possible value attained by $f$ at $x=7$ is

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