Let's solve the derivatives of each transformation given in the table step by step:

(a) $g(x) = f(x) - 2$

• Since $g(x) = f(x) - 2$, the derivative is $g'(x) = f'(x)$. This is because the derivative of a constant ($-2$) is zero.
• Thus, the values for $g'(x)$ will be the same as those for $f'(x)$:
  • $g'(-2) = f'(-2) = 6$
  • $g'(-1) = f'(-1) = \frac{1}{3}$
  • $g'(0) = f'(0) = -\frac{2}{3}$
  • $g'(1) = f'(1) = -1$
  • $g'(2) = f'(2) = -4$
  • $g'(3) = f'(3) = -6$

(b) $h(x) = 2f(x)$

• For $h(x) = 2f(x)$, the derivative is $h'(x) = 2f'(x)$.
• Multiply each $f'(x)$ value by 2:
  • $h'(-2) = 2 \cdot 6 = 12$
  • $h'(-1) = 2 \cdot \frac{1}{3} = \frac{2}{3}$
  • $h'(0) = 2 \cdot -\frac{2}{3} = -\frac{4}{3}$
  • $h'(1) = 2 \cdot -1 = -2$
  • $h'(2) = 2 \cdot -4 = -8$
  • $h'(3) = 2 \cdot -6 = -12$

(c) $r(x) = f(-3x)$

• For $r(x) = f(-3x)$, use the chain rule: $r'(x) = f'(-3x) \cdot (-3)$.
• To find $r'(x)$ at specific values, substitute $-3x$ into $f'(x)$:
  • $r'(-2) = f'(6) \cdot (-3) = \text{undefined}$ (because $f'(6)$ is not given)
  • $r'(-1) = f'(3) \cdot (-3) = -6 \cdot (-3) = 18$
  • $r'(0) = f'(0) \cdot (-3) = -\frac{2}{3} \cdot (-3) = 2$
  • $r'(1) = f'(-3) \cdot (-3) = \text{undefined}$ (because $f'(-3)$ is not given)
  • $r'(2) = f'(-6) \cdot (-3) = \text{undefined}$ (because $f'(-6)$ is not given)
  • $r'(3) = f'(-9) \cdot (-3) = \text{undefined}$ (because $f'(-9)$ is not given)

(d) $s(x) = f(x + 2)$

• For $s(x) = f(x + 2)$, use the chain rule: $s'(x) = f'(x + 2)$.
• Shift the input values by 2:
  • $s'(-2) = f'(0) = -\frac{2}{3}$
  • $s'(-1) = f'(1) = -1$
  • $s'(0) = f'(2) = -4$
  • $s'(1) = f'(3) = -6$
  • $s'(2) = f'(4) = \text{undefined}$ (because $f'(4)$ is not given)
  • $s'(3) = f'(5) = \text{undefined}$ (because $f'(5)$ is not given)

Would you like a more detailed explanation of any part of these calculations?

Related Questions:

1. What is the chain rule, and how is it applied in derivative transformations?
2. How does multiplying a function by a constant affect its derivative?
3. Why does shifting the input of a function affect its derivative in a particular way?
4. What happens to the derivative when composing functions like $f(g(x))$?
5. How do undefined points in the derivative table affect function analysis?

Tip:

When using the chain rule, always remember to multiply the derivative of the inner function by the derivative of the outer function for accurate results!