GATE CS 2021 Set 2 Q25 — Calculus: Limits, Continuity, Differentiability

We can solve this problem using Lagrange's Mean Value Theorem (LMVT).

According to LMVT, if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one point $c \in (a, b)$ such that:

Given:

• Interval: $[-3, 3]$, so $a = -3$ and $b = 3$.
• $f(-3) = 7$
• $f'(x) \le 2$ for all $x \in (-3, 3)$

Applying LMVT on $[-3, 3]$:

Since $f'(x) \le 2$ for all $x$ in the interval, we must have $f'(c) \le 2$:

Therefore, the maximum possible value (at most) of $f(3)$ is 19.