Math Problem Statement

let f(x)=x21\sqrt{x^2-1} And g(x)=14x2\frac{1}{\sqrt{4-x^2}} Find (fg)(x)

Solution

Given the functions f(x)=x21f(x) = \sqrt{x^2 - 1} and g(x)=14x2g(x) = \frac{1}{\sqrt{4 - x^2}}, we need to find the product function (fg)(x)(fg)(x).

The product of two functions f(x)f(x) and g(x)g(x) is defined as:

(fg)(x)=f(x)g(x)(fg)(x) = f(x) \cdot g(x)

Substituting the given functions:

(fg)(x)=x2114x2(fg)(x) = \sqrt{x^2 - 1} \cdot \frac{1}{\sqrt{4 - x^2}}

This simplifies to:

(fg)(x)=x214x2(fg)(x) = \frac{\sqrt{x^2 - 1}}{\sqrt{4 - x^2}}

Therefore, the function (fg)(x)(fg)(x) is:

(fg)(x)=x214x2(fg)(x) = \frac{\sqrt{x^2 - 1}}{\sqrt{4 - x^2}}

Domain Considerations:

  • For f(x)=x21f(x) = \sqrt{x^2 - 1} to be defined, x210x^2 - 1 \geq 0, which means x1x \leq -1 or x1x \geq 1.
  • For g(x)=14x2g(x) = \frac{1}{\sqrt{4 - x^2}} to be defined, 4x2>04 - x^2 > 0, which means 2<x<2-2 < x < 2.

The domain of (fg)(x)(fg)(x) is the intersection of the domains of f(x)f(x) and g(x)g(x): [ \text{Domain of } (fg)(x) = [-2, -1] \cup [1, 2] ]

Would you like a detailed explanation of any part of this process?

Related Questions:

  1. What is the domain of f(x)=x21f(x) = \sqrt{x^2 - 1}?
  2. How do you find the domain of g(x)=14x2g(x) = \frac{1}{\sqrt{4 - x^2}}?
  3. Can you graph the functions f(x)f(x) and g(x)g(x)?
  4. How would you find the zeros of the function (fg)(x)(fg)(x)?
  5. How does the domain restriction affect the behavior of (fg)(x)(fg)(x)?

Tip:

When combining functions, always pay close attention to the domains to avoid including values that make the functions undefined.