Restrict Domain of Function

[nycmathguy]

Section 1.9
Question 72

Can you work out 72 for me to use as reference notes to work out a few more on my own?

 

[MathLover1]

72.
f(x)=abs(x-5)
restrict the domain so that f(x) is one-to-one and has an inverse

How To:

Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.

Without any restriction to its domain, the graph of f(x)=abs(x-5) would fail the horizontal line test because a horizontal line will intersect at it more than once.

If we are going to graph this absolute value function without any restriction to its domain, it will look like this. This is the graph of f(x)=abs(x-5) shifted two units to the right.

However, if we apply the restriction of x≤5, the graph of f(x)=abs(x-5) has been modified to be just the left half of the original function. The left half of f(x)=abs(x-5) can be expressed as the line f(x)=-(x-5) for x≤5.

note: line representing 5-x should end at x=5 and does not go in quadrant IV

inverse:
f(x)=-(x-5)

y=-(x-5)
x=-(y-5)
x=-y+5
y=-x+5 or f^-1(x)=5-x

 

[nycmathguy]

72.
f(x)=abs(x-5)
restrict the domain so that f(x) is one-to-one and has an inverse

How To:

Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.

Without any restriction to its domain, the graph of f(x)=abs(x-5) would fail the horizontal line test because a horizontal line will intersect at it more than once.

If we are going to graph this absolute value function without any restriction to its domain, it will look like this. This is the graph of f(x)=abs(x-5) shifted two units to the right.

However, if we apply the restriction of x≤5, the graph of f(x)=abs(x-5) has been modified to be just the left half of the original function. The left half of f(x)=abs(x-5) can be expressed as the line f(x)=-(x-5) for x≤5.

note: line representing 5-x should end at x=5 and does not go in quadrant IV

inverse:
f(x)=-(x-5)

y=-(x-5)
x=-(y-5)
x=-y+5
y=-x+5 or f^-1(x)=5-x

Nicely-done. I will use this reply to answer two more similar questions.

 

[nycmathguy]

72.
f(x)=abs(x-5)
restrict the domain so that f(x) is one-to-one and has an inverse

How To:

Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse.

Without any restriction to its domain, the graph of f(x)=abs(x-5) would fail the horizontal line test because a horizontal line will intersect at it more than once.

If we are going to graph this absolute value function without any restriction to its domain, it will look like this. This is the graph of f(x)=abs(x-5) shifted two units to the right.

However, if we apply the restriction of x≤5, the graph of f(x)=abs(x-5) has been modified to be just the left half of the original function. The left half of f(x)=abs(x-5) can be expressed as the line f(x)=-(x-5) for x≤5.

note: line representing 5-x should end at x=5 and does not go in quadrant IV

inverse:
f(x)=-(x-5)

y=-(x-5)
x=-(y-5)
x=-y+5
y=-x+5 or f^-1(x)=5-x

You said:

This is the graph of f(x)=abs(x-5) shifted two units to the right.

I think the graph is shifted 5 units to the right not two units.

You also said:

The left half of f(x)=abs(x-5) can be expressed as the line f(x)=-(x-5) for x≤5.

How did you get f(x) = -(x - 5)?

 

[MathLover1]

typo, you are right, the graph is shifted 5 units to the right

abs(x-5) is x-5 or -(x-5)

 

[nycmathguy]

typo, you are right, the graph is shifted 5 units to the right

abs(x-5) is x-5 or -(x-5)

Ok. Thanks. I don't feel good. I will text you when time allows.