Chapter 11.7, Problem 49PPE

To determine

To solve the function $f(x)=\frac{(x+2)(x+1)}{x+2}andg(x)=x+1$ and check are the graphs are same.

Answer to Problem 49PPE

Both the graphs are same reason is given on explanation

Explanation of Solution

Given:

Given function

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

Calculation:

In order to solve the equation $f(x)=\frac{(x+2)(x+1)}{x+2}andg(x)=x+1$ ,

  $f(x)=\frac{(x+2)(x+1)}{x+2}$

As $f(x)=\frac{(x+2)(x+1)}{x+2}$ As we know that $f\left(x\right)=y$

So we can say that

  $y=\frac{(x+2)(x+1)}{x+2}$

Considering the values of x and y put the values in function

X	$y=\frac{(x+2)(x+1)}{x+2}$	(x, y)
-1	$y=\frac{(x+2)(x+1)}{x+2}$ y = 0	(-1, 0)
0	$y=\frac{(x+2)(x+1)}{x+2}$ y = 1	(0, 1)

  

And secondly now

  $g(x)=x+1$

As $g(x)=x+1$ and as we know that $g\left(x\right)=y$

So we can say that

  $y=x+1$

x	$y=x+1$	(x, y)
-1	$y=x+1$ y = 0	(-1, 0)
0	$y=x+1$ y = 1	(0, 1)

  

Reason: Here both the graphs are same because both have same value i.e. (-1, 0) in x-axis and (0, 1) in y-axis.

And both are straight line graph making on second quadrant

Thus, the solution is correct.