To solve the given problems, we will evaluate the functions \( f(x) = x^2 - 3x \) and \( g(x) = 8 - x^3 \).
Part 1: (f + g)(4)
First, we calculate \( f(4) \) and \( g(4) \): \[ f(4) = 4^2 - 3(4) = 16 - 12 = 4 \] \[ g(4) = 8 - 4^3 = 8 - 64 = -56 \] Now, we find \( (f + g)(4) \): \[ (f + g)(4) = f(4) + g(4) = 4 + (-56) = 4 - 56 = -52 \]
Answer for Part 1:
\[ (f + g)(4) = -52 \]
Part 2: (g - f)(1)
Next, we calculate \( f(1) \) and \( g(1) \): \[ f(1) = 1^2 - 3(1) = 1 - 3 = -2 \] \[ g(1) = 8 - 1^3 = 8 - 1 = 7 \] Now, we find \( (g - f)(1) \): \[ (g - f)(1) = g(1) - f(1) = 7 - (-2) = 7 + 2 = 9 \]
Answer for Part 2:
\[ (g - f)(1) = 9 \]
Part 3: (f * g)(1)
We already have \( f(1) = -2 \) and \( g(1) = 7 \). Now, we calculate \( (f \cdot g)(1) \): \[ (f \cdot g)(1) = f(1) \cdot g(1) = -2 \cdot 7 = -14 \]
Answer for Part 3:
\[ (f \cdot g)(1) = -14 \]
Part 4: \(\left( \frac{g}{f} \right)(4)\)
We will calculate \( \frac{g(4)}{f(4)} \): From Part 1, we know:
- \( f(4) = 4 \)
- \( g(4) = -56 \)
Now we can calculate: \[ \left( \frac{g}{f} \right)(4) = \frac{g(4)}{f(4)} = \frac{-56}{4} = -14 \]
Answer for Part 4:
\[ \left( \frac{g}{f} \right)(4) = -14 \]
Summary of Answers
a. \( (f + g)(4) = -52 \)
b. \( (g - f)(1) = 9 \)
c. \( (f \cdot g)(1) = -14 \)
d. \( \left( \frac{g}{f} \right)(4) = -14 \)
1 answer