Question 629914
Inverses of of functions are found by swapping the x's and y's. To perform this swap, I like to temporarily replace the function notation with a "y":
f(x) = 2x+1
y = 2x + 1
Swap the x's and y's:
x = 2y + 1   <=== This is the inverse, but not in a desired form
Solve for y:
x - 1 = 2y
Divide by 2 (or multiply by 1/2):
{{{(1/2)x - 1/2 = y}}}
{{{(1/2)x - 1/2 = f^(-1)(x)}}}<br>
Doing the same for g(x):
{{{g(x) = x^2}}}
{{{y = x^2}}}
{{{x = y^2}}}
{{{sqrt(x) = sqrt(y^2)}}}
<u>+</u>{{{sqrt(x) = y}}}
Note: Because of the <u>+</u>, the inverse of g is not a function (i.e. some, in fact all, x's have more than one y value). And since this inverse is not a function, I do not believe it is proper to use function notation, g^-1 on it.<br>
For the compositions it helps if we have a good understanding of what the function's definition is telling you. For example
f(x) = 2x+1
The left side tells us that the input to the function, between the parentheses after the function name, is being called "x". "x" is just a placeholder. It is just being used as a name for the input to the function. We could use <i>any</i> letter here to name the input. f(q) = 2q+1 is the <i>exact same function</i> as f(x)!!<br>
The right side tells us what function f does with its input. It multiplies the input, x, by 2 and then adds 1. Since "x" is just a place holder, function f will take <i>any</i> input, multiply it by 2 and then add 1:
f(7) = 2(7) + 1
f(1002.4) = 2(1002.4) + 1
f(12x-3) = 2(12x-3) + 1
etc.
(fog)(x) is just another way to write f(g(x)). In f's parentheses we have "g(x)". So "g(x)" is the input to f. And what does f do to its input? Answer: it multiplies by 2 and then adds 1!:
f(g(x)) = 2(g(x)) + 1
Since {{{g(x) = x^2}}} we can substitute this in for g(x):
{{{f(g(x)) = 2(g(x)) + 1 = 2(x^2) + 1}}}<br>
Now we find the inverse of fog:
{{{y = 2x^2 + 1}}}
{{{x = 2y^2 + 1}}}
{{{x - 1 = 2y^2}}}
{{{(1/2)x - 1/2) = y^2}}}
<u>+</u>{{{sqrt((1/2)x - 1/2) = y}}}
Again, we get an inverse that is not a function.<br>
For g^-1 o f^-1, we use the two inverses we found back at the start and feed the inverse of f into the inverse of g as its input: From the inverse of g, <u>+</u>{{{sqrt(x) = y}}}, we can see that it finds the positive and negative square roots of its input. So it will do the same to f^1, {{{(1/2)x - 1/2}}}:
<u>+</u>{{{sqrt((1/2)x - 1/2) = y}}}
which matches the inverse of fog!