10 - Lab 10 Rational and Radical fn's.sagews

##Lab 9 Rational and Radical Fn's Here we'll look at domains, asymptotes, and holes.

Combining both function types together first, let's examine the idea of domain. The domain is the set of all inputs that will produce an output. Functions use operations on numbers. We know that up to this point there are two things that we can't do with certain numbers to get results: division by zero and taking an EVEN root of a negative number. Other than that, the sky's the limit. So we have to examine our function for the possibility of either of these two situations occurring. Let's look at a rational function first. We'll use $rat(x)=\frac{x+4}{x^2-5x+6}$. We see that the variable is present in the denominator, so there is the potential for some x-value that could cause $x^2-5x+6$ to equal zero. We simply will let SAGE determine any zeros of the denominator polynomial(there might not be any!) In the block below, I just set the denominator function equal to zero and solve it.

x,y=var('x,y')
solve(x^2-5*x+6==0,x)

[x == 3, x == 2]

Therefore, all Reals except 2 and 3 are members of the domain. Let's look at two more. $\frac{5}{x^2+4}$ and $\frac{x^2-3x+1}{x^3+7x^2-18x-5}$

solve(x^2+4==0,x)
show(solve(x^2+4==0,x))

[x == (-2*I), x == (2*I)]

[$\displaystyle x = \left(-2 i\right)$, $\displaystyle x = \left(2 i\right)$]

The $i$'s indicate the solutions are complex and not real, so the denominator has no real zeros and thus the domain is all reals.

solve(x^3+7*x^2-18*x-5==0,x)
show(solve(x^3+7*x^2-18*x-5==0,x))

[x == -1/2*(1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3)*(I*sqrt(3) + 1) - 1/18*(-103*I*sqrt(3) + 103)/(1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3) - 7/3, x == -1/2*(1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3)*(-I*sqrt(3) + 1) - 1/18*(103*I*sqrt(3) + 103)/(1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3) - 7/3, x == (1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3) + 103/9/(1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3) - 7/3]

[$\displaystyle x = -\frac{1}{2} \, {\left(\frac{1}{18} i \, \sqrt{56729} \sqrt{3} - \frac{1685}{54}\right)}^{\frac{1}{3}} {\left(i \, \sqrt{3} + 1\right)} - \frac{-103 i \, \sqrt{3} + 103}{18 \, {\left(\frac{1}{18} i \, \sqrt{56729} \sqrt{3} - \frac{1685}{54}\right)}^{\frac{1}{3}}} - \frac{7}{3}$, $\displaystyle x = -\frac{1}{2} \, {\left(\frac{1}{18} i \, \sqrt{56729} \sqrt{3} - \frac{1685}{54}\right)}^{\frac{1}{3}} {\left(-i \, \sqrt{3} + 1\right)} - \frac{103 i \, \sqrt{3} + 103}{18 \, {\left(\frac{1}{18} i \, \sqrt{56729} \sqrt{3} - \frac{1685}{54}\right)}^{\frac{1}{3}}} - \frac{7}{3}$, $\displaystyle x = {\left(\frac{1}{18} i \, \sqrt{56729} \sqrt{3} - \frac{1685}{54}\right)}^{\frac{1}{3}} + \frac{103}{9 \, {\left(\frac{1}{18} i \, \sqrt{56729} \sqrt{3} - \frac{1685}{54}\right)}^{\frac{1}{3}}} - \frac{7}{3}$]

Here's where you have to be smarter than the machine!!!! I'm thinking. . . don't odd degree functions have to cross the x-axis at least once? As I examine the 3 zeros, they all seem to be imaginary!! Sometimes SAGE can get a little crazy. I'll use the numerical_approx command to see what the zeros probably are.

numerical_approx(-1/2*(1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3)*(I*sqrt(3) + 1) - 1/18*(-103*I*sqrt(3) + 103)/(1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3) - 7/3,digits=3)
numerical_approx(-1/2*(1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3)*(-I*sqrt(3) + 1) - 1/18*(103*I*sqrt(3) + 103)/(1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3) - 7/3,digits=3)
numerical_approx((1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3) + 103/9/(1/18*I*sqrt(56729)*sqrt(3) - 1685/54)^(1/3) - 7/3,digits=3)

-0.253 -8.95 - 0.000122*I 2.20 + 0.000244*I

It turns out that only the first listed is real, so the best we can do for stating the domain, is to say all real numbers except approximately -.253. Now let's take a look at one radical function. $\sqrt{12-x-x^2}$. We need to determine when the radicand is at least zero.

solve(12-x-x^2>=0,x)
show(solve(12-x-x^2>=0,x))

[[x >= -4, x <= 3]]

[[$\displaystyle x \geq \left(-4\right)$, $\displaystyle x \leq 3$]]

So the small domain is just the interval, $[-4,3]$. You can always graph the denominator(for a rational fn) or the radicand(for a radical fn) to get a handle on what the domain most likely is.

Find the domain of each of the following:

1. $\frac{12x}{x^3-5x^2-2x+10}$

2. $\frac{3x+4}{x^2+7x+2}$

3. $\sqrt{x^3+3x^2-6x-8}$

4. $\frac{x^2-7x+6}{x^3+3x+4}$

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%md
Let's take a look at vertical asymptotes.  We'll investigate the simple rational function, $\frac{3-x}{x^2-x-6}$.  Look at a plot of the function first.  

Let's take a look at vertical asymptotes. We'll investigate the simple rational function, $\frac{3-x}{x^2-x-6}$. Look at a plot of the function first.

plot((3-x)/(x^2-x-6),(x,-4,4),ymin=-10,ymax=10)
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%md
We know that zeros of the denominator are where the vertical asymptotes should be and from the plot it appears that there's one at x=-2.  But the denominator is quadratic!!! There should be another one.  Let's factor the quadratic to find the other zero.

We know that zeros of the denominator are where the vertical asymptotes should be and from the plot it appears that there's one at x=-2. But the denominator is quadratic!!! There should be another one. Let's factor the quadratic to find the other zero.

factor(x^2-x-6)

(x + 2)*(x - 3)

According to this there should be two real zeros and thus two asymptotes. Maybe we need to investigate what makes a curve behave like this near a certain x-value. To do this with the above rational function, I am going to use a programming loop to generate a bunch of values for the function as $x$ gets close to the denominator zeroes(investigating one at a time). We'll squeeze toward -2 first then look at 3 second. Make sure you study the syntax since you'll need to use the loop idea later on.

rat(x)=(3-x)/(x^2-x-6);#give the function a name
#will examine what happens as x gets close to -2
for i in [-1,-2,-3,-4,-5]:
    numerical_approx(rat(-2+10^i),digits=5)
#because I added the small power of 10, I am getting closer to -2 from ABOVE
for i in [-1,-2,-3,-4,-5]:
    numerical_approx(rat(-2-10^i),digits=5)
#approached from BELOW
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%md
Once you execute the above code you will see that the y-values grow astronomically as we close in on x=-2.  Let's see what happens with x=3.  You copy and paste the above code, replace -2 with a 3 and see what happens. . .  

Once you execute the above code you will see that the y-values grow astronomically as we close in on x=-2. Let's see what happens with x=3. You copy and paste the above code, replace -2 with a 3 and see what happens. . .

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%md
The reason this happens in a function like $\frac{3-x}{x^2-x-6}$ is that when the denominator is written in factored form, $(x-3)(x+2)$, we see that the $(x-3)$ factor is the opposite of the numerator factor $(3-x)$.  So if we write the numerator as $-(x-3)$,  these two binomials will cancel and the rational function is just $\frac{-1}{x+2}$.  Of course $\frac{3-x}{x^2-x-6}=\frac{-1}{x+2}$ only when $x\ne 3$!!!!!!!!!  We say that $(3,-.2)$ is a *hole* in the graph.

The reason this happens in a function like $\frac{3-x}{x^2-x-6}$ is that when the denominator is written in factored form, $(x-3)(x+2)$, we see that the $(x-3)$ factor is the opposite of the numerator factor $(3-x)$. So if we write the numerator as $-(x-3)$, these two binomials will cancel and the rational function is just $\frac{-1}{x+2}$. Of course $\frac{3-x}{x^2-x-6}=\frac{-1}{x+2}$ only when $x\ne 3$!!!!!!!!! We say that $(3,-.2)$ is a hole in the graph.

Use SAGE to determine the asymptotes and any holes in the graphs of:

1. $\frac{x^2-1}{(x^2-3x-18)}$

2. $\frac{x^2-4}{x^2-3x-10}$

3. $\frac{3(9-x^2)}{x^2-4x+3}$

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