Analyzing

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describe everything

$\blacktriangleright $ Reasoning: Reasoning is a logical explanation that describes our conclusions, how we arrived at those conclusions, and why we think those conclusions are correct.

Analysis is not a list of conclusions. We are not looking for such a list.

We are looking for the thought process that arrived at the list of conclusions.

Completely analyze

\[ L(x) = \frac {5}{1 + 3 \, e^{-\tfrac {x}{2}}} \]

Note: $L(x)$ is the composition of a rational function with an exponential function.

$L = f \circ g$ where

\[ f(y) = \frac {5}{1 + 3 \, y} \, \text { and } \, g(k) = e^{-\tfrac {k}{2}} \]

$\blacktriangleright $ Domain

$e^{-\tfrac {x}{2}}$ is an exponential function with base $e > 1$. It has a positive leading coefficient and the leading coefficient of the inside linear function is negative.

The range of the exponential function, $e^{-\tfrac {x}{2}}$ is $(0, \infty )$. The range does not include positive negative numbers.

Since, the range of $g(k)$ does not include negative numbers, it does not include $-\frac {1}{3}$, which is the zero of the denominaotr of $f(y)$. The denominator of $L(x)$ cannot equal $0$.

That makes the domain of $L$ all real numbers, $(-\infty , \infty )$.

$\blacktriangleright $ Zeros

$L$ is a quotient, therefore the zeros of $L$ would be the zeros of the numerator (which are not also zeros of the denominator). However, the numerator is a constant function. It has no zeros. Therefore, $L$ has no zeros.

$\blacktriangleright $ Continuity

The denominator of $L$ is a shifted exponential function, so it is continuous everywhere. The numerator is a constant function, so it is continuous everywhere. $L$ is the quotient of two continuous functions. Therefore, it is continuous.

Since the denominator is never equal to $0$, there are no singularities.

$\blacktriangleright $ End-Behavior

The inside function is $e^{-\tfrac {x}{2}}$, which is an exponential function with base $e > 1$. It has a positive leading coefficient and the leading coefficient of the inside linear function is negative.

That means $g(k)$ is a positive decreasing exponential function.

\[ \lim \limits _{k \to -\infty }g(k) = \lim \limits _{k \to -\infty }e^{-\tfrac {k}{2}} = \infty \]

\[ \lim \limits _{k \to \infty }g(k) = \lim \limits _{k \to \infty }e^{-\tfrac {k}{2}} = 0 \]

To get the end-behavior of $L(x)$, we need to see what $f(y)$ is doing when $g(k)$ is behaving as above.

\[ \lim \limits _{x \to -\infty }L(x) = \lim \limits _{y \to \infty }f(y) = \lim \limits _{y \to \infty }\frac {5}{1 + 3 y} = 0 \]

This is because $f(y)$ is a rational function with the degree of the denominator greater than the degree of the numerator.

\[ \lim \limits _{x \to \infty }L(x) = \lim \limits _{y \to 0}f(y) = \frac {5}{1 + 3 (0)} = 5 \]

This is because $f$ is continuous, which means the limit as $y$ approaches $0$ is just the value of the function.

$\blacktriangleright $ Behavior Rate-of-Change Increasing and Decreasing

The inside function, $e^{-\tfrac {x}{2}}$, is an exponential function with base $e > 1$. It has a positive leading coefficient and the leading coefficient of the inside linear function is negative.

$e^{-\tfrac {x}{2}}$ is a strictly increasing decreasing function

The outside function is a rational function with a constant positive numerator.

The denominator of the fraction is positve and decreasing, which means it is getting smaller.

That makes $f$ an increasing function.

$L$ is a composition of two increasing functions.

$L(x)$ is a strictly increasing decreasing function.

$\blacktriangleright $ Extrema

As a strictly increasing function on $(-\infty , \infty )$, $L$ cannot have global or local maximums or minimums.

$\blacktriangleright $ Range

$L$ is a continuous function, never negative, and increasing. In addition, we nkow that

\[ \lim \limits _{x \to -\infty }L(x) = \lim \limits _{y \to \infty }f(y) = \lim \limits _{y \to \infty }\frac {5}{1 + 3 y} = 0 \]

\[ \lim \limits _{x \to \infty }L(x) = \lim \limits _{y \to 0}f(y) = \frac {5}{1 + 3 (0)} = 5 \]

The range of $L$ is $(0, 5)$.

$\blacktriangleright $ A Graph

The end-behavior tells us that the graph has horizontal asymptotes.

with Calculus

A Peek Ahead...

Calculus will give us a formula for the derivative.

\[ L'(x) = \frac {-15}{2} \cdot \frac {e^{-\tfrac {x}{2}}}{\left (1+3 e^{-\tfrac {x}{2}}\right )^2} \]

This derivative has no zeros, which tells us that $L(x)$ has no critical numbers. $L'(x)$ is always positive, which tells us that $L(x)$ is always increasing.

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More Examples of Analyzing Functions

2025-08-07 13:40:27