Math Problem Statement

For the given functions f and g, complete parts (a)-(h). For parts (a)-(d), also find the domain. f(x) = 2x + 3; g(x) = 9x - 2.

Solution

Let's solve the given questions step by step.

We have two functions:

  • f(x)=2x+3f(x) = 2x + 3
  • g(x)=9x2g(x) = 9x - 2

(a) Find (f+g)(x)(f + g)(x)

(f+g)(x)=f(x)+g(x)=(2x+3)+(9x2)(f + g)(x) = f(x) + g(x) = (2x + 3) + (9x - 2) Simplifying: (f+g)(x)=2x+9x+32=11x+1(f + g)(x) = 2x + 9x + 3 - 2 = 11x + 1

Now, we are asked to find the domain of f+gf + g. Since both f(x)f(x) and g(x)g(x) are linear functions, which are defined for all real numbers, the domain of f+gf + g is all real numbers: Domain of f+g: (,)\text{Domain of } f + g: \ (-\infty, \infty) So, the correct choice is B: The domain is {x | x is any real number}.

(b) Find (fg)(x)(f - g)(x)

(fg)(x)=f(x)g(x)=(2x+3)(9x2)(f - g)(x) = f(x) - g(x) = (2x + 3) - (9x - 2) Simplifying: (fg)(x)=2x9x+3+2=7x+5(f - g)(x) = 2x - 9x + 3 + 2 = -7x + 5

Similarly, the domain of fgf - g is also all real numbers since both f(x)f(x) and g(x)g(x) are linear functions: Domain of fg: (,)\text{Domain of } f - g: \ (-\infty, \infty) So, again the correct choice is B: The domain is {x | x is any real number}.

Would you like any further details or clarification on this?

Here are 5 additional related questions to explore:

  1. What is the sum of f(x)f(x) and g(x)g(x) evaluated at x=2x = 2?
  2. What is the difference between f(x)f(x) and g(x)g(x) when x=1x = -1?
  3. What would be the domain of f(x)g(x)f(x) \cdot g(x)?
  4. What is the domain of f(x)g(x)\frac{f(x)}{g(x)}?
  5. How would you find the composite function f(g(x))f(g(x))?

Tip: When dealing with linear functions, the domain is typically all real numbers unless there's division by zero or square roots of negative numbers involved.

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Math Problem Analysis

Mathematical Concepts

Linear Functions
Algebra
Function Operations

Formulas

f(x) + g(x) = (2x + 3) + (9x - 2)
f(x) - g(x) = (2x + 3) - (9x - 2)

Theorems

Linear functions have a domain of all real numbers

Suitable Grade Level

Grades 8-10