Composition (?)/Inverse of Functions...HELP

[Drano365]

First of all, thanks in advance for any help...I think I have an idea of what to do, but then again...

The problem the teacher gave -as far as I can tell - has not been over in class - and the textbook as well doesn't, although I may just be seriously missing something.

*********Consider the function f(x)=3x+2. Find a function g that makes each equation true.
a. f(g(x))=x
b. g(f(x))=x
c. f(g(x))=x^2
d. g(f(x))=x^2
e. g(f(x))=f(x)
f. f(g(x)=f(x)
*********

Again, thanks in advance for any help received.

 

[stapel]

Consider the function f(x)=3x+2. Find a function g that makes each equation true.
a. f(g(x))=x
b. g(f(x))=x
c. f(g(x))=x^2
d. g(f(x))=x^2
e. g(f(x))=f(x)
f. f(g(x)=f(x)

Hm... This is an interesting exercise. I don't think I've seen this sort of problem before.

I'd start by trying simple stuff.

a) Ending up with just "x" means that we're looking for the inverse function. So follow the usual procedure to find the inverse.

b) Same as for (a).

c) Let's try something really simple. To get a squared variable, considering that we'll be applying a linear function (namely, f) to something or other, we'll need to have a squared variable inside. So let's try a generic quadratic: ax^2 + bx + c.

f(ax^2 + bx + c) = 3(ax^2 + bx + c) + 2 = 3ax^2 + 3bx + 3c + 2

This will equal x^2 if 3ax^2 = 1x^2, 3bx = 0x, and 3c + 2 = 0. Solve each of these equations for the values of a, b, and c which will give the desired results.

d) This time, we're plugging 3x + 2 into something else. We want to get an x^2. Since we're inputting something linear, we'll need to square. So let's try the generic quadratic again:

g(3x + 2) = a(3x + 2)^2 + b(3x + 2) + c = a(9x^2 + 12x + 4) + 3bx + 2b + c = 9ax^2 + (12a + 3b)x + (4a + 2b + c)

We need 9a = 1, 12a + 3b = 0, and 4a + 2b + c = 0. Solve to find g.

e) I'd go simple for this one. There's an obvious choice for g(x) which will always spit out whatever you input.

f) I think we can use the same function as in (e).

 

[HallsofIvy]

First of all, thanks in advance for any help...I think I have an idea of what to do, but then again...

The problem the teacher gave -as far as I can tell - has not been over in class - and the textbook as well doesn't, although I may just be seriously missing something.

*********Consider the function f(x)=3x+2. Find a function g that makes each equation true.
a. f(g(x))=x
b. g(f(x))=x
c. f(g(x))=x^2
d. g(f(x))=x^2
e. g(f(x))=f(x)
f. f(g(x)=f(x)
*********

Again, thanks in advance for any help received.

Did you not even try? Given that f(x)= 3x+ 2, then f(g(x)= 3g+ and g(f(x)= g(3x+ 2).

a. f(g(x))= x. Solve 3g+ 2= x for g.
c. f(g(x))= x^2. Solve 3g+ 2= x^2 for g.
f. f(g(x))= f. Solve 3g+ 2= 3x+ 2.

The others, for an unknown g, is a little harder. Saying that g(f(x))= g(3x+ 2)= x means that g is the inverse function to f(x)= 3x+ 2 so this is the same as (a). Saying that g(f(x))= x^2 is a little harder. For (d), since f is linear, it is a reasonable assumption that g(x) is quadratic. If g(x)= ax^2+ bx+ c means that g(f(x))= af^2+ bf+ g= a(3x+ 2)^2+ b(3x+ 2)+ c= 9ax^2+ 12ax+ 4a+ 3bx+ 2b+ c= 9ax^2+ (12a+ 3b)x+ (4a+ 2b+ c). So

d. g(f(x))= 9ax^2+ (12a+ 3b)x+ (4a+ 2b+ c)= x^2 for all x. That mean that the coefficients must be the same: 9a= 1, 12a+ 3b= 0, 4a+ 2b+ c= 0.

g. g(f(x)= 9ax^2+ (12a+ 3b)x+ (4a+ 2b+ c)= f(x)= 3x+ 2. That means that 9a= 0, 12a+ 3b= 3, 4a+ 2b+ c= 2.