Functions Problems

Hi
Could someone please help with these function problems - it would be greatly appreciated. I'm having some trouble.

f(x) = 2x + 5 g(x) = 3x+1 h(x) = 1 + 2/x
h o h(x) = ? I put 1 + 2/(1 + 2/x) but that's not the answer
h o f(x) = ? I put 1 + 2/(2x+5) but seems I'm wrong
h o g(x) = ? I put 1 + 2/(3x+1) but not right also

I'm ok with one's like f o g(x) but having trouble with the h ones.

Inverse of f(x) = 1/(2x-5)

Given f(x) = 2x + 5 g(x) = 3x + 1 h(x) = 1 + 2/x express
f o h^-1(x)
g^-1 o f^-1(x)
f o g^-1(x)

Inverse of f(x)=x^2 + 2x + 5 I got f^-1(x) = sqrt((x-5)/2) but that's wrong

Any assistance or advice would be great. Thanks.

Maybe I'm just sleepy, but the first three look fine. Are you entering this into some program for grading? If so, are you entering it correctly, or does it want you to somehow simplify?

As for the inverses, you said "Given f(x) = 2x + 5" and then later "f(x)=x^2 + 2x + 5"--which one is it? Where did the x^2 come from?

Not entering them into a program - just maths questions.
The solutions from the book are:
h o h(x) = 3x + 2 / (x + 2)
h o f(x) = 2x + 7 / (2x + 5)
h o g(x) = 3(x+1)/(3x + 1)
Maybe they simplfy them but I'm not good at that - if anyone can show how they got that it would help a lot.

And on your last bit they are different questions. Sorry if that's confusing.

Your answers are right, they are just simplifying, except that their simplification doesn't simplify things. Here is a simple one followed by the hardest one.

As an example, here's h o f(x):

h o f(x) = 1 + 2/(2x+5)
h o f(x) = (2x+5)/(2x+5) + 2/(2x+5) because (2x+5)/(2x+5) = 1
h o f(x) = ((2x+5) + 2)/(2x+5)
h o f(x) = (2x+7)/(2x+5)

It's the same for h 0 h(x), but the simplification is more complicated:

h(x) = 1 + 2/x
h o h(x) = 1 + 2/(1+2/x)
h o h(x) = (1+2/x)/(1+2/x) + 2/(1+2/x) because (1+2/x)/(1+2/x) = 1
h o h(x) = ((1+2/x)+2)/(1+2/x)
h o h(x) = (2/x+3)/(1+2/x)
h o h(x) = (2/x + 3x/x)/(x/x + 2/x)
h o h(x) = ((3x+2)/x)/((x+2)/x)
now multiply by x/x, which is the same as multiplying by 1, but it gives:
h o h(x) = (x(3x+2)/x)/(x(x+2)/x)
h o h(x) = (3x+2)/(x+2)

Quote:

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Originally Posted by math magic

...
Given f(x) = 2x + 5 g(x) = 3x + 1 h(x) = 1 + 2/x express
f o h^-1(x)
g^-1 o f^-1(x)
f o g^-1(x)
...

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For the inverse functions, remember that f o f^-1 = x i.e. applying f followed by its inverse must yield the original value. Therefore:

f^-1(x) = (x - 5) / 2
g^-1(x) = (x + 1) / 3
h^-1(x) = 2 / (x - 1)

Applying these:

f o h^-1 = 4 / (x - 1) + 5
g^-1 o f^-1 = [(x - 5) / 2 + 1] / 3
f o g^-1 = 2*[(x + 5) / 3] + 5

Thank you to every one for their help. I understand it a lot better now.

krtxmrtz or anyone can you just go through how to get the h^-1(x) which you got as 2 / (x - 1).

Also finally just need help in how to get the inverse of f(x) = 1 / (2x - 5), f(x) = x^2 + 2x + 5, f(x) = sqrt(4 - x^2) and f(x) = (2x - 5)^3 + 3. Was able to get the inverse of most of the questions I required but these gave me some trouble.

The inverse of f(x) = 1 / (2x - 5) can be found by solving for x and plugging in the inverse function:

f(x) = 1/(2x-5) gives (2x-5) = 1/f(x) gives 2x = 1/f(x)+5 gives x = 1/(2f(x))+5/2. Now for x, plug in f^-1(x) to get f^-1(x) = 1/(2f(f^-1(x)) + 5/2 = 1/(2x) + 5/2.

This method should work on any function that you can solve algebraically--so if you can solve for one variable given an equation, you can find the inverse function. The other three inverses... don't look pleasant. Hopefully they simplify, but it's too late in the night for me to try them, and someone will be along shortly to do so anyway if you need help with them :)

Here's one of them, hopefully you can pick up from here for the other 2.

f(x) = x² + 2x + 5 = x² + 2x + 1 + 4 = (x + 1)² + 4

Now you solve for x as j. showed in the previous post:

x = Sqrt(f(x) - 4) -1

Notice f(x) has 2 inverses due to the fact that the square root can be positive or negative, therefore:

f(x)^-1 = Sqrt(f(f(x)^-1 - 4) - 1 = Sqrt(x - 4) -1

or

f(x)^-1 = -Sqrt(x - 4) - 1