Question: Find $a$ and $b$ such that $f$ is differentiable everywhere.
$$f(x) = \begin{cases} ax^3 & x \leq 8 \\ x^2 + b & x > 8 \end{cases}$$
a = 1/12
b = -64/3
1. **State the problem:** We want to find constants $a$ and $b$ such that the piecewise function
$$f(x) = \begin{cases} ax^3 & x \leq 8 \\ x^2 + b & x > 8 \end{cases}$$
is differentiable everywhere, especially at $x=8$ where the definition changes.
2. **Conditions for differentiability:** For $f$ to be differentiable at $x=8$, it must be continuous and have equal derivatives from the left and right at $x=8$.
3. **Continuity at $x=8$:**
$$\lim_{x \to 8^-} f(x) = \lim_{x \to 8^+} f(x) = f(8)$$
From the left:
$$f(8) = a \cdot 8^3 = 512a$$
From the right:
$$f(8) = 8^2 + b = 64 + b$$
Set equal for continuity:
$$512a = 64 + b \quad \Rightarrow \quad b = 512a - 64$$
4. **Differentiability at $x=8$:**
Derivatives from left and right must be equal:
Left derivative:
$$f'(x) = 3ax^2 \implies f'(8) = 3a \cdot 64 = 192a$$
Right derivative:
$$f'(x) = 2x \implies f'(8) = 16$$
Set equal:
$$192a = 16 \quad \Rightarrow \quad a = \frac{16}{192} = \frac{1}{12}$$
5. **Find $b$ using $a=\frac{1}{12}$:**
$$b = 512 \cdot \frac{1}{12} - 64 = \frac{512}{12} - 64 = \frac{128}{3} - 64 = \frac{128}{3} - \frac{192}{3} = -\frac{64}{3}$$
6. **Final answer:**
$$a = \frac{1}{12}, \quad b = -\frac{64}{3}$$
These values ensure $f$ is continuous and differentiable everywhere, including at $x=8$.