Deleon Haynes
09/04/2023 · Primary School
The functions \( f \) and \( g \) are defined as \( f(x)=4 x-3 \) and \( g(x)=-9 x^{2} \) The domain of \( \left(\frac{f}{g}\right)(x) \) is \( (-\infty, 0) \cup(0, \infty) \). (Type your answer in interval notation.) The domain of \( \left(\frac{g}{f}\right)(x) \) is \( \left(-\infty, \frac{3}{4}\right) \cup\left(\frac{3}{4}, \infty\right) \). (Type your answer in interval notation.) b) \( (f+g)(x)=-9 x^{2}+4 x-3 \) (Simplify your answer.) \( (f-g)(x)=9 x^{2}+4 x-3 \) (Simplify your answer.) \( (f g)(x)=36 x^{3}+27 x^{2} \) (Simplify your answer.)
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Step-by-step Solution
Let's analyze the functions \( f(x) = 4x - 3 \) and \( g(x) = -9x^2 \) and perform the required operations.
### Part a: Domains of \(\frac{f}{g}\) and \(\frac{g}{f}\)
1. **Domain of \(\frac{f}{g}\)**:
The function \(\frac{f}{g}(x) = \frac{4x - 3}{-9x^2}\) is defined wherever \(g(x) \neq 0\).
We find when \(g(x) = -9x^2 = 0\):
\[
-9x^2 = 0 \implies x^2 = 0 \implies x = 0
\]
Thus, the domain of \(\frac{f}{g}\) is all real numbers except \(x = 0\):
\[
\text{Domain of } \frac{f}{g} = (-\infty, 0) \cup (0, \infty)
\]
2. **Domain of \(\frac{g}{f}\)**:
The function \(\frac{g}{f}(x) = \frac{-9x^2}{4x - 3}\) is defined wherever \(f(x) \neq 0\).
We find when \(f(x) = 4x - 3 = 0\):
\[
4x - 3 = 0 \implies 4x = 3 \implies x = \frac{3}{4}
\]
Thus, the domain of \(\frac{g}{f}\) is all real numbers except \(x = \frac{3}{4}\):
\[
\text{Domain of } \frac{g}{f} = \left(-\infty, \frac{3}{4}\right) \cup \left(\frac{3}{4}, \infty\right)
\]
### Part b: Simplifying the operations
1. **Sum: \((f + g)(x)\)**:
\[
(f + g)(x) = f(x) + g(x) = (4x - 3) + (-9x^2) = -9x^2 + 4x - 3
\]
This is already simplified.
2. **Difference: \((f - g)(x)\)**:
\[
(f - g)(x) = f(x) - g(x) = (4x - 3) - (-9x^2) = 4x - 3 + 9x^2 = 9x^2 + 4x - 3
\]
This is also already simplified.
3. **Product: \((f g)(x)\)**:
\[
(f g)(x) = f(x) \cdot g(x) = (4x - 3)(-9x^2)
\]
Distributing:
\[
= -9x^2(4x) + (-9x^2)(-3) = -36x^3 + 27x^2
\]
Thus, the simplified product is:
\[
(f g)(x) = -36x^3 + 27x^2
\]
### Final Answers
- Domain of \(\frac{f}{g}\): \((- \infty, 0) \cup (0, \infty)\)
- Domain of \(\frac{g}{f}\): \(\left(-\infty, \frac{3}{4}\right) \cup \left(\frac{3}{4}, \infty\right)\)
- \((f + g)(x) = -9x^2 + 4x - 3\)
- \((f - g)(x) = 9x^2 + 4x - 3\)
- \((f g)(x) = -36x^3 + 27x^2\)
Quick Answer
Domain of \(\frac{f}{g}\): \((- \infty, 0) \cup (0, \infty)\)
Domain of \(\frac{g}{f}\): \(\left(-\infty, \frac{3}{4}\right) \cup \left(\frac{3}{4}, \infty\right)\)
\((f + g)(x) = -9x^2 + 4x - 3\)
\((f - g)(x) = 9x^2 + 4x - 3\)
\((f g)(x) = -36x^3 + 27x^2\)
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