Inverses of functions and analytic geometry

[GetThroughDiffEq]

Gone through the book twice and these two questions confused me.

1. Determine if the following functions are inverses of each other:

f(x)=-5x+2
g(x)=5x-2

Plugged them and got -5x for both [f o g](x) and [g o f](x), meaning they would be inverses of each other.

Apparently, they are not?

2. Find the perimeter of the triangle having vertices with the given coordinates:

(2, 2), (5, 2), (2, 6)

I then graphed the coordinates into a right triangle.

I understand how to get 4 (y coordinates 2 to 6) and 3 (x coordinates 2 to 5), but am still confused how to calculate the final value (which should be 5) because the perimeter of the triangle should be equal to 12.

 

[Dr.Peterson]

Gone through the book twice and these two questions confused me.

1. Determine if the following functions are inverses of each other:

f(x)=-5x+2
g(x)=5x-2

Plugged them and got -5x for both [f o g](x) and [g o f](x), meaning they would be inverses of each other.

Apparently, they are not?

2. Find the perimeter of the triangle having vertices with the given coordinates:

(2, 2), (5, 2), (2, 6)

I then graphed the coordinates into a right triangle.

I understand how to get 4 (y coordinates 2 to 6) and 3 (x coordinates 2 to 5), but am still confused how to calculate the final value (which should be 5) because the perimeter of the triangle should be equal to 12.

1. f(g(x)) = f(5x - 2) = -5(5x - 2) + 2 = -25x - 8. How did you get -5x?

In any case, if they were inverses, this would be x. So they are not inverses.

2. Use the distance formula (or, equivalently, the Pythagorean Theorem). This is the familiar 3-4-5 triangle.

 

[Steven G]

f(g(x)) = f(5x - 2) = -5(5x - 2) + 2 = -25x + 12

 

[Deleted member 4993]

2. Find the perimeter of the triangle having vertices with the given coordinates:

(2, 2), (5, 2), (2, 6)

I then graphed the coordinates into a right triangle.

I understand how to get 4 (y coordinates 2 to 6) and 3 (x coordinates 2 to 5), but am still confused how to calculate the final value (which should be 5) because the perimeter of the triangle should be equal to 12.

Since you have graphed it - you can see that it is a right-angled triangle (more specifically a 3-4-5 triangle with the length of the hypotenuse being 5). If you have a straight line joining points (A and B) whose co-ordinates are (x_a, y_a) and (x_b, y_b), respectively - then the length of the line (L_AB), joining those points would be:

L²_AB = (x_b- x_a)² + (y_b - y_a)²

 

[pka]

2. Find the perimeter of the triangle having vertices with the given coordinates:
(2, 2), (5, 2), (2, 6)

\(\displaystyle \mathscr{Prm}=\sqrt{(5-2)^2+(2-2)^2}+\sqrt{(2-5)^2+(6-2)^2}+\sqrt{(2-2)^2+(6-2)^2}\)​

 

[JeffM]

Gone through the book twice and these two questions confused me.

1. Determine if the following functions are inverses of each other:

f(x)=-5x+2
g(x)=5x-2

Plugged them and got -5x for both [f o g](x) and [g o f](x), meaning they would be inverses of each other.

Apparently, they are not?

2. Find the perimeter of the triangle having vertices with the given coordinates:

(2, 2), (5, 2), (2, 6)

I then graphed the coordinates into a right triangle.

I understand how to get 4 (y coordinates 2 to 6) and 3 (x coordinates 2 to 5), but am still confused how to calculate the final value (which should be 5) because the perimeter of the triangle should be equal to 12.

Notation is your friend.

[MATH]q = \text {distance between } (2,\ 2) \text { and } (5,\ 2).[/MATH]
[MATH]r = \text {distance between } (5,\ 2) \text { and } (2,\ 6).[/MATH]
[MATH]s = \text {distance between } (2,\ 6) \text { and } (2,\ 2).[/MATH]
[MATH]p = \text {perimter of triangle.}[/MATH]
[MATH]p = q + r + s.[/MATH]
You should know the formula, derived from the Pythagorean Theorem, that gives the distance between two points in a co-ordinate plane, namely

[MATH]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}.[/MATH]
So [MATH]q = \sqrt{(2 - 5)^2 + (2 - 2)^2} = \text {WHAT?}[/MATH]
And [MATH]r = \sqrt{(5 - 2)^2 + (2 - 6)^2} = \text {WHAT?}[/MATH]
And [MATH]q = \sqrt{(2 - 2)^2 + (6 - 2)^2} = \text {WHAT?}[/MATH]
This gets you exactly to where pka started.

You did not need to graph anything. You did not need to realize that the triangle was a right triangle. All you needed was the definition of the perimeter and the distance formula betwen points in a co-ordinate plane.