Question

1. Identify whether the functions shown are rational:
a. f(x) = x / x^2 + 1
b. f(x) = √x / x^2 + 1
c. f(x) = x / x^0.4 + 1
d. f(x) = (x / x^2 + 1)^2
e. f(x) = √2x / ex^2 + √3

Answer 1

Answer:

a) Rational function

b) Not a rational function

c) Not a rational function

d) Rational function

e) Not a rational function

Step-by-step explanation:

A rational function is of the form;

$f(x)=\frac{p(x)}{q(x)} ,\:q(x)\ne 0$ where $p(x)$ and $q(x)$ are polynomial functions.

For option a) $f(x)=\frac{x}{x^2+1}$, we have both the numerator $p(x)=x$ and the denominator $q(x)=x^2+1$ are polynomial functions, hence $f(x)=\frac{x}{x^2+1}$ is a rational function.

For option b) $f(x)=\frac{\sqrt{x}}{x^2+1}$, $p(x)=\sqrt{x}$ is not a polynomial function hence $f(x)=\frac{\sqrt{x}}{x^2+1}$ is not a rational function.

For option c) we have $f(x)=\frac{x}{x^{0.4}+1}$, $q(x)=x^{0.4}+1$ is not a polynomial function hence $f(x)=\frac{x}{x^{0.4}+1}$ is not a rational function.

For option d) $f(x)=(\frac{x}{x^2+1})^2 =\frac{x^2}{x^4+2x^2+1}$ both $p(x)=x^2$ and $q(x)=x^4+2x^2+1$ are polynomials hence  $f(x)=(\frac{x}{x^2+1})^2$ is a rational function.

For option e) $f(x)=\frac{\sqrt{2x} }{e^{x^2}+\sqrt{3} }$ both the numerator and the denominator are not polynomials hence  $f(x)=\frac{\sqrt{2x} }{e^{x^2}+\sqrt{3} }$  is not a rational function.