Question

Let $f : [-1,3] \to \mathbb{R}$ be a continuous function such that $f$ is differentiable on $(-1,3)$, $|f'(x)| \le \dfrac{3}{2}$ for all $x \in (-1,3)$, $f(-1) = 1$ and $f(3) = 7$. Then $f(1)$ equals .................
 

Hint:

When a bound on $|f'(x)|$ is given, it represents the maximum slope the function can attain. Use this to estimate intermediate function values using linearity and the Mean Value Theorem.

Answer 1

Correct Answer: 4

Step 1: Apply the Mean Value Theorem (MVT).
For a differentiable function on $(-1,3)$ and continuous on $[-1,3]$, there exists $c \in (-1,3)$ such that \[ f'(c) = \frac{f(3) - f(-1)}{3 - (-1)} = \frac{7 - 1}{4} = \frac{3}{2}. \] This is the maximum possible derivative allowed by the condition $|f'(x)| \le \dfrac{3}{2}$.

Step 2: Find the possible range for $f(1)$.
By the Lipschitz condition $|f'(x)| \le \dfrac{3}{2}$, we have \[ |f(x_2) - f(x_1)| \le \frac{3}{2} |x_2 - x_1|. \] Using $f(-1) = 1$ and $f(3) = 7$, the total change in $f$ from $-1$ to $3$ is 6.

Step 3: Estimate $f(1)$.
The interval $[-1,3]$ has length 4. To maintain the derivative bound, the rate of change per 2-unit interval from $-1$ to $1$ and $1$ to $3$ should balance around the maximum slope $\dfrac{3}{2}$. Hence, \[ f(1) - f(-1) = 2 \times \frac{3}{2} = 3 \implies f(1) = 1 + 3 = 4. \]

Step 4: Verification.
The slope between $(1,3)$ is $\dfrac{7 - 4}{2} = 1.5 = \dfrac{3}{2}$, consistent with the given bound. \[ \boxed{f(1) = 4.} \]