Math Problem Statement

Find (f + g)(-2), (f + g)(1), (f + g)(-4), and the domain of f, g, and f + g, based on the given functions f(x) = x + 1 and g(x) = sqrt(x + 3).

Solution

Let's break down the problem and solve the questions based on the provided graph and functions:

Given Functions:

  1. f(x)=x+1f(x) = x + 1
  2. g(x)=x+3g(x) = \sqrt{x + 3}
  3. (f+g)(x)=f(x)+g(x)=x+1+x+3(f + g)(x) = f(x) + g(x) = x + 1 + \sqrt{x + 3}

Part (a): Find (f+g)(2)(f + g)(-2) and (f+g)(1)(f + g)(1)

Step 1: Calculate (f+g)(2)(f + g)(-2)

We substitute x=2x = -2 into the equation for (f+g)(x)(f + g)(x):

(f+g)(2)=(2+1)+2+3=1+1=1+1=0(f + g)(-2) = (-2 + 1) + \sqrt{-2 + 3} = -1 + \sqrt{1} = -1 + 1 = 0

So, (f+g)(2)=0(f + g)(-2) = 0.

Step 2: Calculate (f+g)(1)(f + g)(1)

We substitute x=1x = 1 into the equation for (f+g)(x)(f + g)(x):

(f+g)(1)=(1+1)+1+3=2+4=2+2=4(f + g)(1) = (1 + 1) + \sqrt{1 + 3} = 2 + \sqrt{4} = 2 + 2 = 4

So, (f+g)(1)=4(f + g)(1) = 4.

Part (b): Can you find (f+g)(4)(f + g)(-4)?

To calculate (f+g)(4)(f + g)(-4), substitute x=4x = -4 into the equation:

(f+g)(4)=(4+1)+4+3=3+1(f + g)(-4) = (-4 + 1) + \sqrt{-4 + 3} = -3 + \sqrt{-1}

Since the square root of a negative number is not a real number, (f+g)(4)(f + g)(-4) is undefined in the real number system.

Part (c): Find the domain of ff, gg, and f+gf + g

Domain of f(x)=x+1f(x) = x + 1

The function f(x)=x+1f(x) = x + 1 is a linear function, which is defined for all real numbers. So, the domain of ff is: Domain of f:(,)\text{Domain of } f: (-\infty, \infty)

Domain of g(x)=x+3g(x) = \sqrt{x + 3}

For the square root function g(x)=x+3g(x) = \sqrt{x + 3}, the argument inside the square root must be non-negative: x+30x3x + 3 \geq 0 \quad \Rightarrow \quad x \geq -3 So, the domain of gg is: Domain of g:[3,)\text{Domain of } g: [-3, \infty)

Domain of (f+g)(x)(f + g)(x)

Since (f+g)(x)=f(x)+g(x)(f + g)(x) = f(x) + g(x), the domain of f+gf + g is the intersection of the domains of ff and gg. Since f(x)f(x) is defined for all xx, the domain of f+gf + g is the same as the domain of gg: Domain of (f+g):[3,)\text{Domain of } (f + g): [-3, \infty)

Summary of Results:

  • (f+g)(2)=0(f + g)(-2) = 0
  • (f+g)(1)=4(f + g)(1) = 4
  • (f+g)(4)(f + g)(-4) is undefined.
  • Domain of ff: (,)(-\infty, \infty)
  • Domain of gg: [3,)[-3, \infty)
  • Domain of f+gf + g: [3,)[-3, \infty)

Would you like further clarification on any part? Here are some related questions to consider:

  1. What happens if you add another function h(x)h(x) to f+gf + g?
  2. Can you find (fg)(x)(f \cdot g)(x) for x=1x = 1?
  3. What is the range of g(x)g(x)?
  4. Can f(x)f(x) and g(x)g(x) intersect, and if so, where?
  5. How would you solve f(x)=g(x)f(x) = g(x)?

Tip: Always check the domain of each function when combining them to avoid undefined expressions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Functions
Algebra
Square Roots
Domains

Formulas

f(x) = x + 1
g(x) = sqrt(x + 3)
(f + g)(x) = f(x) + g(x) = x + 1 + sqrt(x + 3)

Theorems

Properties of square roots
Function domain intersection

Suitable Grade Level

Grades 9-11