The best way to analyze a rationa function is with its factored form.

Analyze R(x)

\[ R(x) = \frac {(x+5)(x-3)}{x+2}, \text { with its natural domain } \]

We want to say as much as we can exactly. Then, we’ll turn to the graph for approximations.

\(\blacktriangleright \) Domain: We are given that the domain is the natural domain. Therefore, we need to exclude any zero of the denominator. \((-\infty , -2) \cup (-2, \infty )\).

\(\blacktriangleright \) Continuity: All rational functions are continuous on their domain. So, \(R\) has no discontinuities and just one singularity.

\(\blacktriangleright \) End-Behavior:

\(R\) is a rational function. The degree of the numerator is \(2\), which is greater than than the degree of the denominator. Therefore, \(R\) is unbounded as \(x\) tends to \(-\infty \) or \(\infty \). We just have to figure out the sign.

When \(x\) is very large and negative, we have

  • \(x+5 < 0\)
  • \(x-3 < 0\)
  • \(x+2 < 0\)

That makes \(R < 0\)

\[ \lim \limits _{x \to -\infty } R(x) = -\infty \]

When \(x\) is very large and positive, then all three factors are positive and \(R\) is positive.

\[ \lim \limits _{x \to \infty } R(x) = \infty \]

\(\blacktriangleright \) Zeros:

We are given the factored form. So, we can see that the zeros are \(-5\) and \(3\).

Both zeros and the singularity have a multiplicity of \(1\), which is odd. Therefore, \(R(x)\) changes signs over each one.

\(\blacktriangleright \) Behavior Around the Singularity:

When \(x\) is close to \(-2\), but less than \(-2\), then we have

  • \(x+5 ~ 3 > 0\)
  • \(x-3 ~ -5 < 0\)
  • \(x+2 ~ 0 < 0\)

The numerator will be around \(-15\) and the denominator will be heading to \(0\) through negative numbers. \(R\) will be positive and unbounded.

\[ \lim \limits _{x \to -2^-} R(x) = \infty \]

When \(x\) is close to \(-2\), but greater than \(-2\), then we have

  • \(x+5 ~ 3 > 0\)
  • \(x-3 ~ -5 < 0\)
  • \(x+2 ~ 0 > 0\)

The numerator will be around \(-15\) and the denominator will be heading to \(0\) through positive numbers. \(R\) will be negative and unbounded.

\[ \lim \limits _{x \to -2^+} R(x) = -\infty \]

\(\blacktriangleright \) Range:

\(B\) is continuous on \((-\infty , -2)\) and \(\lim \limits _{x \to -\infty } R(x) = -\infty \) and \(\lim \limits _{x \to -2^-} R(x) = \infty \). This tells us that the range is \((-\infty , \infty )\).

So far, all of our algebraic reasoning agrees with the graph.

\(\blacktriangleright \) Behavior:

Now, we want to figure out where \(R\) is increasing and decreasing and then use that information to identify any local maximums or minimums.

\(\blacktriangleright \) Our algebra cannot do that. We’ll need Calculus to do that.

At this point, we can only turn to the graph for approximations.

The graph suggests that \(R\) is always increasing and there are no local maximums or minimums.

with Calculus...

Calculus will give us some new algebra tools by which we can obtain a formula for a derivative.

\[ B'(x) \frac {x^2 + 4x + 19}{(x+2)^2} \]
  • Where \(B'(x) < 0\), \(B\) is increasing.
  • Where \(B'(x) > 0\), \(B\) is decreasing.
  • Where \(B'(x) = 0\) or DNE identifies critical numbers.

The quadratic formula tells us that the zeros of the numerator are not real numbers.

\[ \frac {-4 \pm \sqrt {16 - 4 (1) (19)}}{2} \]

Therefore, the numerator is always the same sign. It never changes sign. Since, the numerator equals \(19\) when \(x=0\), we know that the numerator is always positive.

The denominator is a square, so it is never negative.

This shows that \(B'(x) > 0\) on the whole domain and \(B(x)\) is always increasing.

2025-01-09 20:20:57