Checking an inverse: W = 50 + 2.3(h - 60), h = 72; H = (W-50)/(2.3) + 60

[Johnw321]

Hello everyone, I am am doing some studying and cant figure out what this question is asking me.Verify that the function given in number 1 and your function from number 2 are inverses, by showing that https://blackboard.cpcc.edu/bbcswebdav/pid-7430549-dt-content-rid-46181232_1/xid-46181232_1 and that https://blackboard.cpcc.edu/bbcswebdav/pid-7430549-dt-content-rid-46181228_1/xid-46181228_1.

Now my first function was W = 50 + 2.3(h - 60), with h being 72. I solved it and got 77.6 for W. It then asked me to rewrite it to make it a function for H so I rewrote it as H=(W-50)/(2.3)+60. Now when I plug in W(77.6) I do end up with the original h(72). Is it basically asking me to show that process or is it asking me something else. I'm confusded;(

 

[tkhunny]

Not bad. Showing a single value, however, does not prove that something is an inverse.

If f(x) and g(x) are inverses, then f(g(x)) = x and g(f(x)) = x. One must substitute the ENTIRE function expression.

Do that.

 

[Johnw321]

but?

So would I write for (w*h)(w)=w ..... W=50+2.3(77.6-50)/(2.3)+60-60 times 50+2.3(72-60) it's not making sense to me

Not bad. Showing a single value, however, does not prove that something is an inverse.

If f(x) and g(x) are inverses, then f(g(x)) = x and g(f(x)) = x. One must substitute the ENTIRE function expression.

Do that.

 

[tkhunny]

Function Notation would help.

W(h) means "W" is a function of "h".

To find an inverse, we want h(W), or "h" is a function of "W".

If they are inverses, they will express the same relationship

W = 50 + 2.3(h - 60)

This could also be stated as W(h) = 50 + 2.3(h - 60)
And the function notation works nicely.

W(0) = 50 + 2.3(0 - 60) = 50 + 2.3(-60) = 50 - 138 = -88
W(60) = 50 + 2.3(60 - 60) = 50
W(Frog) = 50 + 2.3(Frog - 60)

We need the right potential inverse. You have that. There are several ways to write it. Your version is fine.

h = (W-50)/2.3 + 60
OR, in Function Notation:
h(W) = (W-50)/2.3 + 60

Note: Keep track of your variables. "H" and "h" are not the same thing. I stayed with the lower case.

And again, the function notation works nicely:

h(0) = (0-50)/2.3 + 60 = 57.826
h(50) = (W-50)/2.3 + 60 = 60
h(Pittsburgh) = (Pittsburgh-50)/2.3 + 60

Your challenge is to produce two things:

W(h(W)) = ??
h(W(h)) = ??

Just use the function notation EXACTLY like it is shown above. Whatever it is inside those parentheses, that is what you substitute for the variable, wherever it appears.

 

[mmm4444bot]

… by showing that

and that

.

It looks like something is missing, after each word 'that'.

 

[stapel]

Hello everyone, I am am doing some studying and cant figure out what this question is asking me.Verify that the function given in number 1 and your function from number 2 are inverses, by showing that https://blackboard.cpcc.edu/bbcswebdav/pid-7430549-dt-content-rid-46181232_1/xid-46181232_1 and that https://blackboard.cpcc.edu/bbcswebdav/pid-7430549-dt-content-rid-46181228_1/xid-46181228_1.

Now my first function was W = 50 + 2.3(h - 60), with h being 72. I solved it and got 77.6 for W. It then asked me to rewrite it to make it a function for H so I rewrote it as H=(W-50)/(2.3)+60. Now when I plug in W(77.6) I do end up with the original h(72). Is it basically asking me to show that process or is it asking me something else. I'm confusded;(

The images to which you link are password-protected, so we are unable to view them.

Kindly please reply with the typed-out text of the exercise, along with its instructions. For instance, guessing from your mention of some of your work done, a full explanation may have been along the lines of the following:

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. . .the first two questions were:

. . .1. (something or other about) the following function:

. . . . .\(\displaystyle W(h)\, =\, 50\, +\, 2.3\, (h\, -\, 60)\)

. . .2. Find the inverse of the function in Exercise (1).

. . .my working for the second question was:

. . . . .\(\displaystyle \color{darkgreen}{W\, =\, 50\, +\, 2.3\, (h\, -\, 60)}\)

. . . . .\(\displaystyle \color{darkgreen}{W\, -\, 50\, =\, 2.3\, (h\, -\, 60)}\)

. . . . .\(\displaystyle \color{darkgreen}{\dfrac{W\, -\, 50}{2.3}\, =\, h\, -\, 60}\)

. . . . .\(\displaystyle \color{darkgreen}{\dfrac{W\, -\, 50}{2.3}\, +\, 60\, =\, h}\)

. . .then the third question was:

. . .3. Show algebraically that the function given in (1)
. . .and your function found in (2) are indeed inverses.

. . .to answer this, I picked h = 72....

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However, showing that functions are inverses is very different from showing that one particular point gets mapped correctly between the two functions (which is, I'm guessing, what you were doing by picking a value of "72" for evaluation purposes). Instead, you need to compose the two functions, in each of two ways, as explained in an earlier post.

Please reply with any necessary corrections or clarifications of the above. When you reply, please include a clear listing of your function-composition efforts so far, or else please specify that you're needing lesson instruction on that topic first. Thank you!