Question

Let $f$ be a function which is continuous and differentiable for all $x$. If $f(1) = 1$ and $f'(x) \leq 5$ for all $x$ in $[1, 5]$, then the maximum value of $f(5)$ is

• A. $5$
• B. $20$
• C. $6$
• D. $21$

Hint:

$f(b) \leq f(a) + f'(max)(b-a)$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem uses the Lagrange's Mean Value Theorem (LMVT).
The theorem relates the average rate of change over an interval to the instantaneous derivative.
Step 2: Key Formula or Approach:
LMVT formula on interval $[a, b]$: $\frac{f(b) - f(a)}{b - a} = f'(c)$ for some $c \in (a, b)$.
Here $a = 1, b = 5$. So $\frac{f(5) - f(1)}{5 - 1} \leq \max(f'(x))$.
Step 3: Detailed Explanation:
Given $f'(x) \leq 5$:
\[ \frac{f(5) - f(1)}{4} \leq 5 \] \[ f(5) - f(1) \leq 20 \] Substitute $f(1) = 1$:
\[ f(5) - 1 \leq 20 \implies f(5) \leq 21 \] The maximum possible value for $f(5)$ is reached when the equality holds.
Step 4: Final Answer:
The maximum value of $f(5)$ is $21$.