Let $f(x)$ be double differentiable function such that $|f''(x)| \leq 5$ $\forall\, x \in [0, 4]$ and $f$ takes its largest value at an interior point of this interval. Then the value of $|f'(0)| + |f'(4)|$ can be:
Step-by-Step Solution
Key Concept: The core idea involves using Fermat's Theorem, which states that if a differentiable function attains its largest value at an interior point $c$, then its derivative at that point must be zero, i.e., $f'(c)=0$. This is then combined with the Mean Value Theorem (or its integral form) applied to $f'$ to establish a bound on the change in $f'$ given a bound on $f''$, specifically $|f'(b)-f'(a)| \leq M|b-a|$ if $|f''(x)| \leq M$.
Step 1: Since $f$ takes its largest value at an interior point $c \in (0,4)$, by Fermat's theorem $f'(c) = 0$.
Step 2: Apply the Mean Value Theorem (or use the bound on $f''$). Since $|f''(x)| \leq 5$, we have $|f'(x) - f'(y)| \leq 5|x-y|$ for all $x, y \in [0,4]$.
Step 3: Bound $|f'(0)|$. Since $f'(c) = 0$ and $c \in (0,4)$:
$$|f'(0)| = |f'(0) - f'(c)| \leq 5|0 - c| = 5c < 5 \cdot 4 = 20$$
Step 4: Bound $|f'(4)|$. Similarly:
$$|f'(4)| = |f'(4) - f'(c)| \leq 5|4 - c| = 5(4-c) < 20$$
Step 5: Add the bounds:
$$|f'(0)| + |f'(4)| \leq 5c + 5(4-c) = 20$$
So $|f'(0)| + |f'(4)| \leq 20$.
Step 6: Since the maximum value is 20 (achieved when $c \to 0$ or $c \to 4$, but $c$ is strictly interior), the value is strictly less than 20. Thus $|f'(0)| + |f'(4)| < 20$, meaning possible values are 18 and 19 (and values up to but not including 20). However, the value 20 can be approached arbitrarily closely, so 18, 19, and 20 are all possible values (with 20 being achievable in the limit). The correct options are (a) 18, (b) 19, (c) 20.
Correct Answer: 1, 2, 3