Question

1. Find two functions f(x) and g(x) such that f[g(x)] = x but g[f(x)] does not equal x.2. State the vertical, horizontal asymptotes and zeros of the rational function, f(x) = {x^2+3x+2}/{x^2+5x+4}.Why is there no zero at x = –1?3. Give an example and explain why a polynomial can have fewer x-intercepts than its number of roots.

Answer 1

Answer:

1. $f(x)=x^{2}$ and $g(x)=-\sqrt{x}$

2. Vertical Asymptote: x=-4

    Horizontal Asymptote: y=1

    Zeros: (-2,0)

    Because the graph will have a hole at x=-1. This is, the graph is not defined for that particular x-value.

3.  $f(x)=x^{2}+4$

It has two complex roots, so there is no place in the graph where it crosses the x-axis.

Step-by-step explanation:

1. If we find the composite function f[g(x)] with the functions:

$f(x)=x^{2}$ and g(x)=-x, we will get the following:

$f[g(x)]=(-\sqrt{x})^{2}$

which simplifies to

f[g(x)]=x

on the other hand, if we find g[f(x)] we will get the following:

$g[f(x)]=-\sqrt{x^2}$

which simplifies to:

g[f(x)]=-x

2.

Before finding the asymptotes, it's a good idea to factor both the numerator and denominator of the function, so we get:

$f(x)=\frac{x^{2}+3x+2}{x^{2}+5x+4}$

$f(x)=\frac{(x+2)(x+1)}{(x+4)(x+1)}$

we can now see that this function can be simplified. It is important to simplify this function so we don't confuse vertical asymptotes with holes in the graph, so when simplifying we get the following:

$f(x)=\frac{x+2}{x+4}$

now we can set the denominator equal to zero.

x+4=0

and solve for x:

x=-4 this is our vertical asymptote.

For the horizontal asymptote, there is a rule that tells us that if the polynomials on the numerator and denominator have the same degree (the same greatest power), then the horizontal asymptote is found by dividing the lead coefficient of the polynomial on the numerator by the lead coefficient on the denominator. In this case the lead coefficients are 1 and 1 so the horizontal asymptote is:

$y=\frac{1}{1}$

so the horizontal asymptote is at:

y=1.

The zeros of the function are located at the x-values that will turn the numerator of the simplified fraction equal to zero, so we get:

x+2=0

x=-2

so the only zero will be located at (-2,0)

x=-1 is not a zero because the original function is not defined for x=-1, there is a hole in the graph at this x-value.

Take a look at the graph of the function (See attached picture)

3.

This will happen when there are no real zeros. Take for example the function:

$f(x)=x^{2}+4$

In this case this problem has 2 non-real zeros:

$x^{2}+4=0$

$x^{2}=-4$

$x=\sqrt{-4}$

so:

x=2i   or   x=-2i

since they are complex numbers they cannot be graphed on a coordinate axis, so the graph will not cross the x-axis. See attached picture.