How do you find the zeros of the function $f\left(x\right)=\frac{{x}^{2}-x-12}{{x}^{2}+2x-35}$?

Jamar Hays
2022-09-13
Answered

How do you find the zeros of the function $f\left(x\right)=\frac{{x}^{2}-x-12}{{x}^{2}+2x-35}$?

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asked 2021-02-25

True or False. The graph of a rational function may intersect a horizontal asymptote.

asked 2021-06-16

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals.

asked 2022-06-14

I was bored in my Algebra 2 class and wanted to try to illustrate the tan(x) function as an infinite summation of rational functions.

I recalled that the solution to a rational function's numerator is equivalent to the roots of the function. The denominator's solution is the vertical asymptote(s). Since tan(x) has infinite roots and asymptotes, if I expressed it as a rational function it'd be infinitely long. My teacher suggested just using an infinite summation of rational functions, so I did that instead.

The first thing I did was set up the numerator and denominator so that they could each equal the corresponding roots and asymptotes; since it's (sorta? kinda? maybe) an even function, I just used variations of $x\pm \pi $ for both sides. Given that all roots of tan(x) are whole multiples of pi and all asymptotes are odd multiples of pi/2, I came up with the following equation:

$\sum _{n=-\mathrm{\infty}}^{\mathrm{\infty}}\frac{x\pm n\pi}{x\pm \frac{2n-1)\pi}{2}}$

Where n is any integer from negative infinity to infinity, my idea is that the end product of this is tan(x). Is this even remotely correct or did I leave something out? Thanks!

(I do recognize that the horizontal asymptote is apparently 1 with this equation, but when I plot a section of this in desmos it seems to pass through y=1 anyway. There might be something going on that I could simplify in terms of rewriting an "x^2" somewhere, or maybe I'm just trippin.)

I recalled that the solution to a rational function's numerator is equivalent to the roots of the function. The denominator's solution is the vertical asymptote(s). Since tan(x) has infinite roots and asymptotes, if I expressed it as a rational function it'd be infinitely long. My teacher suggested just using an infinite summation of rational functions, so I did that instead.

The first thing I did was set up the numerator and denominator so that they could each equal the corresponding roots and asymptotes; since it's (sorta? kinda? maybe) an even function, I just used variations of $x\pm \pi $ for both sides. Given that all roots of tan(x) are whole multiples of pi and all asymptotes are odd multiples of pi/2, I came up with the following equation:

$\sum _{n=-\mathrm{\infty}}^{\mathrm{\infty}}\frac{x\pm n\pi}{x\pm \frac{2n-1)\pi}{2}}$

Where n is any integer from negative infinity to infinity, my idea is that the end product of this is tan(x). Is this even remotely correct or did I leave something out? Thanks!

(I do recognize that the horizontal asymptote is apparently 1 with this equation, but when I plot a section of this in desmos it seems to pass through y=1 anyway. There might be something going on that I could simplify in terms of rewriting an "x^2" somewhere, or maybe I'm just trippin.)

asked 2022-07-12

If $k$ is a field in which $1+1\ne 0$, prove that $\sqrt{1-{x}^{2}}$ is not a rational function. Hint. Mimic the classical proof that $\sqrt{2}$ is irrational

It seems rather odd to discuss the square root function in the context of an arbitrary field, but here goes nothing!

Proof:

Suppose that $\sqrt{1-{x}^{2}}$ is a rational function in $k(x)$. Then there exist polynomials $p(x)$ and $q(x)\ne 0$ that are relatively prime with $\sqrt{1-{x}^{2}}=\frac{p(x)}{q(x)}$; clearly we may take $p(x)\ne 0$. Then $1-{x}^{2}=\frac{p(x{)}^{2}}{q(x{)}^{2}}$ or $q(x{)}^{2}(1-{x}^{2})=p(x{)}^{2}$, which says $q|{p}^{2}$, and since $q|{q}^{2}$, we can infer $q|{p}^{2}$. Moreover, since $p|{p}^{2}$ and $(p,q)=1$, then $pq|{p}^{2}$ or $q|p$, which contradicts the fact that the polynomials are relatively prime. Hence $\sqrt{1-{x}^{2}}$ cannot be a rational function.

How does this sound? Aside from the problem of discussing –$\sqrt{\text{}\text{}}$ in the context of an arbitrary field, I am worried about not using the fact that $1+1\ne 0$, at least not explicitly. Where exactly is this assumed used, if at all?

EDIT:

Suppose that $q(x)=1$, and let $f(x)={a}_{n}{x}^{n}+...{a}_{1}x+{a}_{0}$. Then $1-{x}^{2}=f(x{)}^{2}$. If $x=0$, $1={a}_{0}^{2}$ and therefore ${a}_{0}$ must be a unit. Letting $x=1$ we get $0=({a}_{n}+...+{a}_{1}+{a}_{0}{)}^{2}$; and letting $x=-1$ we get $0=(-{a}_{n}-...-{a}_{1}+{a}_{0}{)}^{2}$. Since we are working in a field, there can be no nonzero nilpotent which means that ${a}_{n}+...+{a}_{1}+{a}_{0}$ and $-{a}_{n}-...-{a}_{1}+{a}_{0}$ are both zero. Adding the two equations together yields $2{a}_{0}=0$, and since $1+1\ne 0$, ${a}_{0}=0$ which contradicts the fact that ${a}_{0}$ is a unit.

How does this sound?

Another attemtpt:

If $f(x{)}^{2}=1-{x}^{2}$, then $2=\mathrm{deg}({f}^{2})=2\mathrm{deg}(f)$ implies $\mathrm{deg}(f)=1$. Yet if $x=-1$, we get $f(-1{)}^{2}=1-1=0$ and therefore $f(-1)=0$ since there are no nonzero nilpotent elements in a field; similarly, $f(1)=0$. Since $1+1\ne 0$, $f$ has two distinct roots in $k$ yet is only a 1-st degree polynomial, a contradiction.

It seems rather odd to discuss the square root function in the context of an arbitrary field, but here goes nothing!

Proof:

Suppose that $\sqrt{1-{x}^{2}}$ is a rational function in $k(x)$. Then there exist polynomials $p(x)$ and $q(x)\ne 0$ that are relatively prime with $\sqrt{1-{x}^{2}}=\frac{p(x)}{q(x)}$; clearly we may take $p(x)\ne 0$. Then $1-{x}^{2}=\frac{p(x{)}^{2}}{q(x{)}^{2}}$ or $q(x{)}^{2}(1-{x}^{2})=p(x{)}^{2}$, which says $q|{p}^{2}$, and since $q|{q}^{2}$, we can infer $q|{p}^{2}$. Moreover, since $p|{p}^{2}$ and $(p,q)=1$, then $pq|{p}^{2}$ or $q|p$, which contradicts the fact that the polynomials are relatively prime. Hence $\sqrt{1-{x}^{2}}$ cannot be a rational function.

How does this sound? Aside from the problem of discussing –$\sqrt{\text{}\text{}}$ in the context of an arbitrary field, I am worried about not using the fact that $1+1\ne 0$, at least not explicitly. Where exactly is this assumed used, if at all?

EDIT:

Suppose that $q(x)=1$, and let $f(x)={a}_{n}{x}^{n}+...{a}_{1}x+{a}_{0}$. Then $1-{x}^{2}=f(x{)}^{2}$. If $x=0$, $1={a}_{0}^{2}$ and therefore ${a}_{0}$ must be a unit. Letting $x=1$ we get $0=({a}_{n}+...+{a}_{1}+{a}_{0}{)}^{2}$; and letting $x=-1$ we get $0=(-{a}_{n}-...-{a}_{1}+{a}_{0}{)}^{2}$. Since we are working in a field, there can be no nonzero nilpotent which means that ${a}_{n}+...+{a}_{1}+{a}_{0}$ and $-{a}_{n}-...-{a}_{1}+{a}_{0}$ are both zero. Adding the two equations together yields $2{a}_{0}=0$, and since $1+1\ne 0$, ${a}_{0}=0$ which contradicts the fact that ${a}_{0}$ is a unit.

How does this sound?

Another attemtpt:

If $f(x{)}^{2}=1-{x}^{2}$, then $2=\mathrm{deg}({f}^{2})=2\mathrm{deg}(f)$ implies $\mathrm{deg}(f)=1$. Yet if $x=-1$, we get $f(-1{)}^{2}=1-1=0$ and therefore $f(-1)=0$ since there are no nonzero nilpotent elements in a field; similarly, $f(1)=0$. Since $1+1\ne 0$, $f$ has two distinct roots in $k$ yet is only a 1-st degree polynomial, a contradiction.

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Could any one give me a hint how to show Every rational function which is holomorphic on every point of Riemann Sphere( ${\mathbb{C}}_{\mathrm{\infty}}$) must be constant?(with out applying Maximum Modulas Theorem). Thanks.

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