Help with Functions/Inverse Function/Domains/Ranges/Expressions

Everytime i tried to post my text below it kept crashing your webserver interal 500. So i just took a sreenshot and cropped it below.

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myristate said:

Everytime i tried to post my text below it kept crashing your webserver interal 500.

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Yeah... We all have problems with that, from time to time. We still can't figure out quite why. Sorry 'bout that.

myristate said:

Q1: For the function f(x) = 1/(x+1):

a) State the domain of f(x)

My answer: x + 1 != 0, so x != -1

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Correct.

myristate said:

b) Find an expression for the inverse function f^-1(x)

My answer:

f(x) = 1/(x + 1)
y = 1/(x + 1) <= rename
x = 1/(y + 1) <= swap variables
x(y + 1) = 1 <= multiply thru by y + 1
y + 1 = 1/x <= divide thru by x

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You were good to here. But I don't understand what you did next:

myristate said:

y - 1 = 1/x - 1

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How did the "plus one" on the left-hand side become a "minus one"? (If you'd subtracted 1 from each side, then you'd have had just "y=", so something else must be going on here.)

myristate said:

c) Find the range of the original function f(x)

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Hint: The range of f(x) will be the domain of f^-1(x).

myristate said:

Q2: Functions f and g are defined by the equations:

. . . . .f(t) = t² + 3. . . . .g(t) = 2t

a) Find an expression for f(g(t)), and simplify the expression as far as possible.

My work:

t(2t)² + 3
t(4t) + 3

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To learn about function composition, try here. Once you have studied the lesson and worked through the examples, return to this exercise. In particular, note that:

. . . . .f(g(t))

...means that you plug the expression for g(t) into the formula for f(t) in the place of the variable, not multiplied against the variable.