<p>Let <em>f</em>(<em>x</em>) be defined as follows:<br>\[f(x) = \begin{cases} b - x & -2 \le x < -1 \\ a(x+1)^3 + (x+1) + 2b & -1 \le x < 0 \end{cases}\]<br>If <em>f</em> is differentiable at <em>x</em> = −1, find the value of <em>b</em> − <em>a</em>.</p>
Step-by-Step Solution
Key Concept: For f to be differentiable at x = -1, it must be continuous at x = -1 AND the left derivative must equal the right derivative at that point. This requires two conditions: f(-1⁻) = f(-1⁺) and f'(-1⁻) = f'(-1⁺).
Step 1: Apply Continuity at $x = -1$
For $f(x)$ to be differentiable at $x=-1$, it must first be continuous at $x=-1$.
The left-hand limit as $x \to -1$:
$$ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (b - x) = b - (-1) = b + 1 $$
The right-hand limit as $x \to -1$:
$$ \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (ax^2 - 3x + 5) = a(-1)^2 - 3(-1) + 5 = a + 3 + 5 = a + 8 $$
For continuity, the limits must be equal:
$$ b + 1 = a + 8 $$
$$ b - a = 7 \quad (1) $$
Step 2: Apply Differentiability at $x = -1$
For $f(x)$ to be differentiable at $x=-1$, the left-hand derivative must equal the right-hand derivative.
The derivative of the left piece is:
$$ \frac{d}{dx}(b - x) = -1 $$
So, the left-hand derivative at $x=-1$ is $f'(-1^-) = -1$.
The derivative of the right piece is:
$$ \frac{d}{dx}(ax^2 - 3x + 5) = 2ax - 3 $$
So, the right-hand derivative at $x=-1$ is $f'(-1^+) = 2a(-1) - 3 = -2a - 3$.
For differentiability, the derivatives must be equal:
$$ -1 = -2a - 3 $$
$$ 2a = -2 $$
$$ a = -1 \quad (2) $$
Step 3: Find $b - a$
Substitute the value of $a$ from equation (2) into equation (1):
$$ b - (-1) = 7 $$
$$ b + 1 = 7 $$
$$ b = 6 $$
Therefore, the value of $b - a$ is:
$$ b - a = 6 - (-1) = 6 + 1 = 7 $$
Correct Answer: 1.5