f(x) = x^2-5x+4/x-4 g(x) = x-1
Let's make a table of values for the two functions above.
x
f(x)=x^2-5x+4/x-4
g(x) = x-1
0
f( 0)=0^2-5( 0)+4/0-4=-1
g( 0) = 0-1 = -1
1
f( 1)=1^2-5( 1)+4/1-4=0
g( 1) = 1-1 = 0
2
f( 2)=2^2-5( 2)+4/2-4=1
g( 2) = 2-1 = 1
3
f( 3)=3^2-5( 3)+4/3-4=2
g( 3) = 3-1 = 2
4
f( 4)=4^2-5( 4)+4/4-4= Undefined
g( 4) = 4-1 = 3
5
f( 5)=5^2-5( 5)+4/5-4=4
g( 5) = 5-1 = 4
6
f( 6)=6^2-5( 6)+4/6-4=5
g( 6) = 6-1 = 5
7
f( 7)=7^2-5( 7)+4/7-4=6
g( 7) = 7-1 = 6
8
f( 8)=8^2-5( 8)+4/8-4=7
g( 8) = 8-1 = 7
9
f( 9)=9^2-5( 9)+4/9-4=8
g( 9) = 9-1 = 8
10
f( 10)=10^2-5( 10)+4/10-4=0
g( 10) = 10-1 = 0
c
From the table we did in Part B we see that g(4)=3 and f(4) is undefined. However, we can tell that f(x)=g(x) for all other values of x. That is, x=4 is the only point where the functions are different.
f(x) = x^2-5x+4/x-4 g(x)=x-1
The reason why f(4) is undefined is that its denominator equals 0 at x=4. In contrast, g(x) is defined for all real values.
d
In this part we will plot the graph of the functions by using a graphing calculator.
f(x) = x^2-5x+4/x-4 g(x) = x-1
We begin by pushing the Y= button and typing the corresponding equation. We will plot them separately.
Having entered the equation in the calculator, we now push GRAPH to draw it.
At first look, they seem identical. However, let's change the window view. We will check what happens between x=3 and x=5.
We can see that in the graph of the function y=f(x) on the left there is a little hole at point (4,3). On the graph of y=g(x) on the right, there is not. Consequently, both graphs are identical except at x=4.
e
From Part A, we can say that both expressions are equivalent except at x=4.
x^2-5x+4/x-4 = x-1, except atx=4
From Parts B and D we can make a similar conclusion for the functions.
f(x) = g(x), except atx=4
Notice that the point where the functions are different is the one that makes the denominator of f(x) equal to 0.
