Math Problem Statement

f(x)=x^2-9 for x ∈ R, x<0 g(x)=10-x/2 for x ∈ R, x>6 findthe range of fg(x)

Solution

To find the range of f(g(x))f(g(x)), we will analyze the composition of f(x)=x29f(x) = x^2 - 9 and g(x)=10x2g(x) = 10 - \frac{x}{2} under the given constraints.


Step 1: Determine the domain of g(x)g(x)

Given g(x)=10x2g(x) = 10 - \frac{x}{2} for x>6x > 6, the output of g(x)g(x) is:

g(x)(1062,)=(7,)g(x) \in (10 - \frac{6}{2}, \infty) = (7, \infty)

So g(x)>7g(x) > 7.


Step 2: Apply g(x)g(x) to f(x)f(x)

We now evaluate f(g(x))f(g(x)). Substituting g(x)=10x2g(x) = 10 - \frac{x}{2} into f(x)=x29f(x) = x^2 - 9, we get:

f(g(x))=(g(x))29f(g(x)) = (g(x))^2 - 9 f(g(x))=(10x2)29f(g(x)) = \left(10 - \frac{x}{2}\right)^2 - 9

Simplify (10x2)2\left(10 - \frac{x}{2}\right)^2:

(10x2)2=10010x+x24\left(10 - \frac{x}{2}\right)^2 = 100 - 10x + \frac{x^2}{4} f(g(x))=10010x+x249f(g(x)) = 100 - 10x + \frac{x^2}{4} - 9 f(g(x))=9110x+x24f(g(x)) = 91 - 10x + \frac{x^2}{4}


Step 3: Range of f(g(x))f(g(x))

The domain of g(x)g(x) implies g(x)>7g(x) > 7, but f(x)=x29f(x) = x^2 - 9 is defined only for x<0x < 0. Thus, for g(x)>7g(x) > 7, we need g(x)g(x) to fall in the part of f(x)f(x) corresponding to x<0x < 0.

  1. From g(x)>7g(x) > 7, solve x29x^2 - 9 for x<0x < 0.

Finalizing the Range of f(g(x))f(g(x)):

We are tasked with finding the range of f(g(x))=x29f(g(x)) = x^2 - 9, where g(x)>7g(x) > 7, and f(x)=x29f(x) = x^2 - 9 is defined only for x<0x < 0.


Step 1: Analyze f(x)=x29f(x) = x^2 - 9 for x<0x < 0

The function f(x)=x29f(x) = x^2 - 9 is a parabola that opens upwards, symmetric about x=0x = 0, and restricted to x<0x < 0. For x<0x < 0, x2x^2 decreases as xx approaches 00 from the left. Thus:

Range of f(x)=(9,)for x<0.\text{Range of } f(x) = (-9, \infty) \quad \text{for } x < 0.


Step 2: Analyze g(x)g(x) for x>6x > 6

The function g(x)=10x2g(x) = 10 - \frac{x}{2} decreases linearly for x>6x > 6. The domain implies g(x)>7g(x) > 7, meaning:

g(x)(7,).g(x) \in (7, \infty).


Step 3: Composition f(g(x))f(g(x))

For g(x)>7g(x) > 7, substitute g(x)g(x) into f(x)=x29f(x) = x^2 - 9. Since f(x)f(x) is defined for x<0x < 0, we must check where g(x)g(x) maps back into the valid input domain of f(x)f(x).

a. Solve g(x)<0g(x) < 0 for g(x)>7g(x) > 7:

Let g(x)=yg(x) = y. For f(y)=y29f(y) = y^2 - 9, yy must also be y<0y < 0.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Function Composition
Quadratic Functions
Piecewise Functions

Formulas

f(x) = x^2 - 9
g(x) = 10 - x/2
f(g(x)) = (10 - x/2)^2 - 9

Theorems

Properties of Quadratic Functions
Domain and Range of Functions

Suitable Grade Level

Grades 11-12