Please help with some math questions?

Question

Let f(x) = 2x + 1, g(x) = x2 – 1 and h(x) = , find:

a. (f ◦ g) (x) b. (h ◦ g)(x) c. (g ◦ f)(x)

d. f -1 - the inverse of f e. h-1

Please i just need some explanation of how to do it or an example of a similar question.

Answer 1

Veeeeeeeeeeeery good question.

Well,

a) 2( 2x - 1 ) + 1

b) h(x) not given so there is no answer for this question...

c) ( 2x + 1 ) 2 -1

d) set x = 2y + 1 (inverse function: swap the x and y)

so it ends up as: y = ( x - 1 ) / 2

e) h(x) not given again.

Answer 2

I prefer the notation f(g(x)) to the "little circle", as it makes it much clearer to me what's going on.

a)  f(g(x)) = f(x^2 - 1) = 2(x^2 - 1) + 1 = 2x^2 - 1

b)  This is going to be hard to do without knowing what h(x) is.

c)  g(f(x)) = g(2x + 1) = (2x + 1)^2 - 1 = 4x^2 - 4x + 1 - 1 = 4x^2 - 4x

d)  f^(-1)(x)  is the function that "undoes" f(x)  You want to undo multiplying by 2 and adding 1, so you'll subtract 1 and divide by 2.  A little more analytic way to get there:  let f(x) = y.

y = 2x + 1

y - 1 = 2x

x = (y - 1)/2

So, f^(-1)(x) = (x - 1)/2

You can check this.  If it's right, f(f^-1(x)) = f^-1(f(x)) = x.  In other words, if you compose the function and inverse function in either order, you must end up with just x.

e)  And we run into the same "what's h" problem.

Answer 3

for a, just write f equation, write it again, but without the x and put the g equations every time there was supposed to be x in the f equation, like 2(x2-1) +1 and solve, do this for b and c, and i forgot how to do d sorrry, good luck