One-to-One Functions; g & h defined.

[jynny]

Hello there!
I'm brand new to these forums, and I'm in desperate need of help.

I believe i understand the first two parts of this problem. But i can't for the life of me understand the last part!
Please help! If I've made a mistake with the first two parts, please don't hesitate to correct me.
I'm reviewing for an Exam tomorrow and want to be able to do this type of problem on my own.

Thank you so much!

 

[Deleted member 4993]

View attachment 10433


View attachment 10433

h(x) = 3x+ 2

x = 3y + 2

x - 2 = 3y

y = (x-2)/3

h^-1(x) = (x-2)/3

 

[Dr.Peterson]

View attachment 10433

Hello there!
I'm brand new to these forums, and I'm in desperate need of help.

I believe i understand the first two parts of this problem. But i can't for the life of me understand the last part!
Please help! If I've made a mistake with the first two parts, please don't hesitate to correct me.
I'm reviewing for an Exam tomorrow and want to be able to do this type of problem on my own.

Thank you so much!

You are correct on the first question: g^-1(1) = -5, because g(-5) = 1.

You are wrong about the second. If it were true that h^-1(x) = 3x+2, then h(3x+2) would be x, but it is not: 3(3x+2)-2 = 9x+4.
Did you follow the procedure you were taught?

As for the last question, do you know what composition of function is? (h^-1 ∘ h)(2) means h-1(h(2)). What do you get when you apply h to 2, then apply h^-1 to the result? Note that this question is a check of your answer to the second question.

 

[JeffM]

We define the IDENTITY FUNCTION as \(\displaystyle I(x) = x \text { for any x.}\)

Assuming \(\displaystyle f(x)\) and \(\displaystyle f^{-1}(x)\)

are both defined when x = a, then \(\displaystyle f(f^{-1}(a)) = I(a) = f^{-1}(f(a)).\)

That is the whole idea of inverses. Combine them appropriately, and you get an identity.

The additive inverse of x is - x, and x + (- x) = 0. Zero is the additive identity because a + 0 = a.

The multiplicative inverse of x is (1/x) unless x = 0, which has no multiplicative inverse.
x * (1/x) = 1. And 1 is the multiplicative identity because a * 1 = a.

\(\displaystyle I(a) = a \text { and } f(f^{-1}(a)) = I(a) = f^{-1}(f(a)) \implies f(f^{-1}(a)) = a = f^{-1}(f(a)).\)

As Dr. P says, you can use this to check your work.

\(\displaystyle f(x) = x^3 + 1 \implies f^{-1}(x) = \sqrt[3]{x - 1}.\)

Now we check.

\(\displaystyle f(f^{-1}(x)) = f(\sqrt[3]{x - 1}) = (\sqrt[3]{x - 1})^3 + 1 = x - 1 + 1 = x.\)

\(\displaystyle f^{-1}(f(x)) = f^{-1}(x^3 + 1) = \sqrt[3]{(x^3 + 1) - 1} = \sqrt[3]{x^3} = x.\)