Pregunta
Review (MATH 8) I.MC. * Rtors \( (1-3) \) : Let \( f(x)=\sqrt{x-3}, g(x)=2 x-3 \) 1) Find \( [g \circ f](7) \). 2) \( \left(\frac{g}{t}\right)(12) \) 3.) \( (f+9)(4) \) 4.) Fud the dmanan o the netatinn \( 2(x+3)=18 \). 3.) If \( f(x)=\sqrt{x+2} \), anthet \( x \) the durian of \( f \). 4.) \( F f(x)=3 x-2 \). Find \( f(t) \). 7.) IF \( f(x)=3 x-2 \). Find the value \( 4 x \) which \( f(x)=4 \). * Reters \( (8-10) \) : Let \( f(x)=x^{2}-1, g(x)=\frac{2}{x+1}, h(x)=2 x \) 8.) Find \( \left[h \circ \frac{f}{g}\right](1) \) 9.) Determine the domain of \( (g \cdot h)(x) \), iv.) Deternmle the doman of \( \left(\frac{h}{f}\right)(x) \). II. IDENTIF C GTION: A. Deteronie the standard forms of the FF Functions. 1.) Constand function: 4.) Identity funation: 3.) Qualuatic function: 4) Saume mod function: -5) Aludate Value function:
Ask by Gibbs Ellis. in the Philippines
Jan 26,2025
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**I. Multiple Choice Problems**
1. \( [g \circ f](7) = 1 \)
2. \( \left(\frac{g}{f}\right)(12) = 7 \)
3. \( (f + 9)(4) = 10 \)
4. \( x = 6 \)
5. Domain of \( f \): \( x \geq -2 \)
6. \( f(t) = 3t - 2 \)
7. \( x = 2 \)
8. \( \left[h \circ \left(\frac{f}{g}\right)\right](1) = 0 \)
9. Domain of \( (g \cdot h)(x) \): All real numbers except \( x = -1 \)
10. Domain of \( \left(\frac{h}{f}\right)(x) \): All real numbers except \( x = 1 \) and \( x = -1 \)
**II. Identification of Standard Function Forms**
1. **Constant Function:** \( f(x) = c \)
2. **Identity Function:** \( f(x) = x \)
3. **Quadratic Function:** \( f(x) = ax^2 + bx + c \)
4. **Absolute Value Function:** \( f(x) = |x| \)
5. **Sum Function:** \( f(x) = x + k \)
6. **Modulus Function:** \( f(x) = |x| \) (Assumed to be the absolute value function)
If you have any specific questions or need further clarification, feel free to ask!
Solución
Sure, I'll help you work through the problems you've listed. Let's tackle them step-by-step.
---
### **I. Multiple Choice Problems**
**Operators (1–10):**
Given:
- \( f(x) = \sqrt{x - 3} \)
- \( g(x) = 2x - 3 \)
- Later functions:
- \( f(x) = x^2 - 1 \)
- \( g(x) = \frac{2}{x + 1} \)
- \( h(x) = 2x \)
---
#### **1) Find \( [g \circ f](7) \)**
**Solution:**
\[ [g \circ f](7) = g(f(7)) \]
First, find \( f(7) \):
\[ f(7) = \sqrt{7 - 3} = \sqrt{4} = 2 \]
Then, find \( g(2) \):
\[ g(2) = 2(2) - 3 = 4 - 3 = 1 \]
**Answer:** \( [g \circ f](7) = 1 \)
---
#### **2) Find \( \left(\frac{g}{f}\right)(12) \)**
**Solution:**
\[ \left(\frac{g}{f}\right)(12) = \frac{g(12)}{f(12)} \]
Calculate \( g(12) \):
\[ g(12) = 2(12) - 3 = 24 - 3 = 21 \]
Calculate \( f(12) \):
\[ f(12) = \sqrt{12 - 3} = \sqrt{9} = 3 \]
\[ \left(\frac{g}{f}\right)(12) = \frac{21}{3} = 7 \]
**Answer:** \( \left(\frac{g}{f}\right)(12) = 7 \)
---
#### **3) Find \( (f + 9)(4) \)**
**Solution:**
Assuming the problem is \( (f + 9)(4) = f(4) + 9 \):
\[ f(4) = \sqrt{4 - 3} = \sqrt{1} = 1 \]
\[ (f + 9)(4) = 1 + 9 = 10 \]
**Answer:** \( (f + 9)(4) = 10 \)
---
#### **4) Solve the equation \( 2(x + 3) = 18 \)**
**Solution:**
\[ 2(x + 3) = 18 \]
Divide both sides by 2:
\[ x + 3 = 9 \]
Subtract 3 from both sides:
\[ x = 6 \]
**Answer:** \( x = 6 \)
---
#### **5) If \( f(x) = \sqrt{x + 2} \), find the domain of \( f \).**
**Solution:**
The expression inside the square root must be non-negative:
\[ x + 2 \geq 0 \]
\[ x \geq -2 \]
**Answer:** The domain of \( f \) is all real numbers \( x \) such that \( x \geq -2 \).
---
#### **6) If \( F f(x) = 3x - 2 \), find \( f(t) \).**
**Solution:**
Assuming \( F \) is a notation error and \( f(x) = 3x - 2 \):
\[ f(t) = 3t - 2 \]
**Answer:** \( f(t) = 3t - 2 \)
---
#### **7) If \( f(x) = 3x - 2 \), find the value of \( x \) for which \( f(x) = 4 \).**
**Solution:**
\[ 3x - 2 = 4 \]
Add 2 to both sides:
\[ 3x = 6 \]
Divide by 3:
\[ x = 2 \]
**Answer:** \( x = 2 \)
---
#### **8) Find \( \left[h \circ \left(\frac{f}{g}\right)\right](1) \)**
Given:
- \( f(x) = x^2 - 1 \)
- \( g(x) = \frac{2}{x + 1} \)
- \( h(x) = 2x \)
**Solution:**
\[ \left(\frac{f}{g}\right)(1) = \frac{f(1)}{g(1)} \]
Calculate \( f(1) \):
\[ f(1) = 1^2 - 1 = 0 \]
Calculate \( g(1) \):
\[ g(1) = \frac{2}{1 + 1} = \frac{2}{2} = 1 \]
\[ \left(\frac{f}{g}\right)(1) = \frac{0}{1} = 0 \]
Now, apply \( h \):
\[ h(0) = 2(0) = 0 \]
**Answer:** \( \left[h \circ \left(\frac{f}{g}\right)\right](1) = 0 \)
---
#### **9) Determine the domain of \( (g \cdot h)(x) \)**
Given:
- \( g(x) = \frac{2}{x + 1} \)
- \( h(x) = 2x \)
**Solution:**
\[ (g \cdot h)(x) = \frac{2}{x + 1} \cdot 2x = \frac{4x}{x + 1} \]
The denominator cannot be zero:
\[ x + 1 \neq 0 \]
\[ x \neq -1 \]
**Answer:** The domain is all real numbers except \( x = -1 \).
---
#### **10) Determine the domain of \( \left(\frac{h}{f}\right)(x) \)**
Given:
- \( h(x) = 2x \)
- \( f(x) = x^2 - 1 \)
**Solution:**
\[ \left(\frac{h}{f}\right)(x) = \frac{2x}{x^2 - 1} \]
The denominator cannot be zero:
\[ x^2 - 1 \neq 0 \]
\[ x \neq 1 \]
\[ x \neq -1 \]
**Answer:** The domain is all real numbers except \( x = 1 \) and \( x = -1 \).
---
### **II. Identification of Standard Function Forms**
**A. Determine the standard forms of the following functions:**
1. **Constant Function:**
\[ f(x) = c \]
- *Description:* A function that always returns the same constant value regardless of the input \( x \).
2. **Identity Function:**
\[ f(x) = x \]
- *Description:* A function that returns the input value as the output.
3. **Quadratic Function:**
\[ f(x) = ax^2 + bx + c \]
- *Description:* A polynomial function of degree 2, where \( a \neq 0 \).
4. **Absolute Value Function:**
\[ f(x) = |x| \]
- *Description:* A function that returns the non-negative value of \( x \).
5. **Additional Functions (Assumed from Context):**
- **Sum Function:**
\[ f(x) = x + k \]
- *Description:* A linear function representing the addition of a constant \( k \) to the input \( x \).
- **Modulus Function:**
Likely referring to periodic functions like sine or cosine, but based on the list, it seems to correspond to the absolute value function already listed.
- **Arithmetic Operations on Functions:**
Functions involving addition, subtraction, multiplication, or division of other functions.
**Note:** The list provided had some typographical errors, so the above interpretations are based on standard function forms commonly studied in mathematics.
---
If you have specific questions about any of these problems or need further clarification on any topic, feel free to ask!
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To find \([g \circ f](7)\), first calculate \(f(7) = \sqrt{7-3} = \sqrt{4} = 2\). Then, plug this into \(g\): \(g(2) = 2(2) - 3 = 4 - 3 = 1\). So, \([g \circ f](7) = 1\)! Now, for \(\left(\frac{g}{t}\right)(12)\), use \(g(12) = 2(12) - 3 = 24 - 3 = 21\). Thus, \(\left(\frac{g}{t}\right)(12) = \frac{21}{t}\). And remember that when combining functions like this, always make sure your variables are clear to avoid any confusion!
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