Trouble with inverse square root function?

Question

I don't think I've done this correctly, but is the inverse for y=SQRTx-3
f^-1(x)=x^2+3? thanka

Answer 1

For y =f(x) = √(x) − 3, {equivalent to your problem, as written} write

x = √(y) and solve for y

√(y) = x + 3

y = (x + 3)²



IF you meant y = √(x − 3), then write

x = √(y − 3) and solve for y



y − 3 = x²



y = x² + 3



Don't underestimate the importance of parentheses!!

Answer 2

You may know of the "switch x and y's" to find inverse functions. Basically, the textbook want's you to understand that composite functions are a proof of inverse functions. f(g(x) is the function f, of g(x). Think of it as the function f, eating the function g(x). In your f function, where ever you see an x, plug in g(x). For example f(x) = (x)^2+(x)^(1/2)+1/(x) and g(x) = (x)^2. Notice the parenthesis I placed around x? Later on in calculus, this becomes a nice little trick to stay organized, just replace everything inside of the parenthesis with g(x). In this example, f(g(x)) = ((x^2)^2) +((x^2)^(1/2))+1/(x^2). These functions however, are not inverse. Yes, these two functions were selected at random, but notice, they are not inverse. The if and only if statement just says "Hey, you may be able to switch x and y around, but if f(g(x)) = x, then it's not okay." An better example of when this is not okay might be trigonometric functions, natural logarithms, and other 'inarguable variables.' Sorry if you already knew about composite functions, you may have. I'm just not sure, different books teach them at different places. I didn't learn that trick well into pre-calculus. Composite functions are cool, and reoccur later on in math! You'll even see some of them in calculus, but they're nothing to be afraid of.

Answer 3

If the original function is y=(SqRt x) - 3, the inverse would be

x = (SqRt y) - 3

x+3 = SqRt y. Now, square both sides.

(x+3)^2 = y or y = x^2 + 6x + 9

Answer 4

f(x) = Sqrt(x - 3)



x = Sqrt(y - 3)

x^2 = y - 3

-x^2 = -y + 3

y = x^2 + 3



f^-1(x) = x^2 + 3

Answer 5

y=sqrt(x-3)

to find f^-1 we switch x and y and solve for y

x=sqrt(y-3)

x^2=y-3

y=x^2+3

f^-1(x)=y=x^2+3

yes you did it correctly. good job :)