Limits, Continuity & Differentiability
Continuity of piecewise functions
Grade 12
<p>Let \(f(x) = x^3 - 9x^2 + 24x - 4\). Let \(g(x)\) be defined as follows:<br>
\[g(x) = \begin{cases} f(x+2); & x < 0 \\ f(x+2); & 0 \leq x < 1 \\ f(x+1); & 1 \leq x \leq 2 \\ 4a - x; & x \geq 2 \end{cases}\]<br>
If \(g(x)\) is continuous for all \(x\), find the value of \(a\).</p>
Step-by-Step Solution
Key Concept: For g(x) to be continuous at x = -2, the left-hand limit (using f(x+2)) must equal the right-hand limit (using ax). Set up the equality: lim(x→-2⁻) f(x+2) = lim(x→-2⁺) ax, then solve for a.
Step 1: For $g(x)$ to be continuous at $x = -2$, the left-hand limit, the right-hand limit, and the function value at $x=-2$ must all be equal.
$$ \lim_{x \to -2^-} g(x) = \lim_{x \to -2^+} g(x) = g(-2) $$
Step 2: Calculate the left-hand limit. For $x < -2$, $g(x) = f(x+2)$.
$$ \lim_{x \to -2^-} g(x) = \lim_{x \to -2^-} f(x+2) $$
Substitute $x = -2$ into $f(x+2)$:
$$ f(-2+2) = f(0) $$
Given $f(x) = x^3 - 9x^2 + 24x - 4$, evaluate $f(0)$:
$$ f(0) = (0)^3 - 9(0)^2 + 24(0) - 4 = -4 $$
Thus, $\lim_{x \to -2^-} g(x) = -4$.
Step 3: Calculate the right-hand limit and the function value at $x=-2$. For $x \ge -2$, $g(x) = ax$.
$$ \lim_{x \to -2^+} g(x) = \lim_{x \to -2^+} (ax) = a(-2) = -2a $$
$$ g(-2) = a(-2) = -2a $$
Step 4: Equate the limits to ensure continuity.
$$ -4 = -2a $$
Solve for $a$:
$$ a = \frac{-4}{-2} $$
$$ a = 2 $$
Correct Answer: (4) 4