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For the following problems, state the domain and range of the given functions: \(f=x^{2}+2 x-3, \qquad g=\ln (x-5), \qquad h=\frac{1}{x+4}\) $$ h \circ f $$

Short Answer

Expert verified
Domain: \( \mathbb{R} \), Range: \((0, \infty)\)

Step by step solution

01

Understand function composition

Function composition involves applying one function to the results of another. The notation \( h \circ f \) means we first apply the function \( f \) to \( x \), and then apply \( h \) to the result, i.e., \( h(f(x)) \).
02

Determine the function of the composition

First, determine the expression for \( f(x) \), which is given as \( f(x) = x^2 + 2x - 3 \). Then compute \( h(f(x)) = \frac{1}{f(x)+4} = \frac{1}{(x^2 + 2x - 3) + 4} = \frac{1}{x^2 + 2x + 1}. \)
03

Simplify the expression

Find the simplified form of \( h(f(x)) \). We simplify \( x^2 + 2x + 1 \) in the denominator as \((x+1)^2\) to get \( h(f(x)) = \frac{1}{(x+1)^2}. \)
04

Determine the domain of the composition

The inside function \( f(x) = x^2 + 2x - 3 \) can take any real number \( x \) (so its domain is all real numbers, \( \mathbb{R} \)). The expression \((x+1)^2\) is never zero, thus \( h(f(x)) \) is defined for all \( x \), meaning the domain of \( h \circ f \) is too all real numbers, \( \mathbb{R} \).
05

Determine the range of the composition

The expression \( \frac{1}{(x+1)^2} \) only outputs positive values (since \((x+1)^2 \geq 0\) and the reciprocal is positive). As \( x \) approaches \( \infty \) or \(-\infty \), \( (x+1)^2 \) becomes larger, thus the value of \( \frac{1}{(x+1)^2} \) approaches \( 0 \) from the positive side. Hence, the range of \( h \circ f \) is \((0, \infty)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain of a Function
When we talk about the domain of a function, we refer to the set of all possible input values (usually represented by \( x \)) for which a function is defined. Think of the domain as the function's "playing field," where it accepts entries and performs calculations. For polynomial functions like \( f(x) = x^2 + 2x - 3 \), the domain is typically all real numbers, \( \mathbb{R} \), meaning you can plug in any real number and the function will still work.

However, things get a bit more complex with functions like logarithmic or rational expressions:
  • Logarithmic functions, like \( g(x) = \ln(x-5) \), require the input to be greater than zero, so the domain starts at 5 and extends to infinity: \((5, \infty)\).
  • Rational functions, such as \( h(x) = \frac{1}{x+4} \), cannot have a denominator of zero. So, in this instance, \( x \) cannot be \(-4\) to avoid making the function undefined.
To determine the domain of the composed function \( h \circ f(x) \), we first look at \( f(x) \). Since \( f(x) \) is a polynomial, its domain is all real numbers. For the composition \( \frac{1}{(x+1)^2} \), none of these values make it undefined because \( (x+1)^2 \) will never be zero. Thus, the domain of \( h \circ f \) is \( \mathbb{R} \) as well.
Range of a Function
The range of a function is the complete set of values that the function can yield as outputs. Think of it as the "target range" for the results of your function's calculations. When analyzing the range, it's pivotal to understand the function's behavior over its domain.

For the function \( h \circ f(x) = \frac{1}{(x+1)^2} \), let's think about the output values. As previously mentioned, because \((x+1)^2\) is a square of a real expression, it is always non-negative and never zero. Consequently, \( \frac{1}{(x+1)^2} \) is always positive.
  • As \( x \) moves further away from -1 in either the positive or negative direction, \( (x+1)^2 \) increases, making \( \frac{1}{(x+1)^2} \) decrease towards zero.
  • As a result, the function does not output zero itself, but can get arbitrarily close to it from the positive side.
Hence, the range of the composite function \( h \circ f \) is \((0, \infty)\), including every positive real number but never reaching or including zero.
Composite Functions
Composite functions involve chaining two or more functions together, like a kind of mathematical assembly line. When we see notation like \( h \circ f \), it means we're plugging the output of the function \( f \) into the function \( h \). This essentially creates a new function with its own unique domain and range.

Consider composing two individual functions:
  • The first function, \( f(x) \), outputs \( x^2 + 2x - 3 \), which acts as the new input for \( h(x) \).
  • The second function, \( h(x) = \frac{1}{x+4} \), is then applied to these results.
  • The result of \( h(f(x)) \), simplified to \( \frac{1}{(x+1)^2} \), becomes our composite function.
Understanding composite functions is crucial, as they underline how changing the order and interaction of functions can impact the overall behavior dramatically. It's a layered approach that opens up new possibilities and analyses for solving complex mathematical problems. By examining the domains and ranges of each component and the resulting composite, you can fully grasp and predict the behavior of these functions in any given context.

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