Question: x -3 0 3 6
f(x) -5 4 1 7
f'(x) -1 2 -2 4
The table above gives values of a twice-differentiable function $f$ and its first derivative $f'$ for selected values of $x$. Let $g$ be the function defined by $g(x) = f(x^2 - x)$.
(a) What is the value of $g'(3)$ ?
(b) It is known that $g''(0) = -1$. What is the value of $f''(0)$ ?
1. **State the problem:**
We have a function $g(x) = f(x^2 - x)$ where $f$ is twice differentiable. We know values of $f$, $f'$, and $g''(0)$, and we want to find $f''(0)$.
2. **Recall the chain rule for the first derivative:**
$$g'(x) = f'(x^2 - x) \cdot (2x - 1)$$
3. **Find the second derivative $g''(x)$ using the product and chain rules:**
$$g''(x) = \frac{d}{dx} \left[ f'(x^2 - x) \cdot (2x - 1) \right]$$
Apply the product rule:
$$g''(x) = f''(x^2 - x) \cdot \frac{d}{dx}(x^2 - x) \cdot (2x - 1) + f'(x^2 - x) \cdot \frac{d}{dx}(2x - 1)$$
4. **Calculate the derivatives inside:**
$$\frac{d}{dx}(x^2 - x) = 2x - 1$$
$$\frac{d}{dx}(2x - 1) = 2$$
5. **Substitute these back into the expression:**
$$g''(x) = f''(x^2 - x) \cdot (2x - 1) \cdot (2x - 1) + f'(x^2 - x) \cdot 2$$
Or equivalently:
$$g''(x) = f''(x^2 - x) (2x - 1)^2 + 2 f'(x^2 - x)$$
6. **Evaluate $g''(0)$:**
Substitute $x=0$:
$$g''(0) = f''(0^2 - 0) (2 \cdot 0 - 1)^2 + 2 f'(0^2 - 0)$$
$$g''(0) = f''(0) \cdot (-1)^2 + 2 f'(0) = f''(0) + 2 f'(0)$$
7. **Use the given values:**
From the table, $f'(0) = 2$ and $g''(0) = -1$ (given).
8. **Set up the equation and solve for $f''(0)$:**
$$-1 = f''(0) + 2 \times 2$$
$$-1 = f''(0) + 4$$
$$f''(0) = -1 - 4 = -5$$
**Final answer:**
$$\boxed{-5}$$