[MOVED] Find inverse of y = x^2 - 3, graph, consider domains

[usman77]

For the following functions

Equation --> y=x^2-3

a).Find Inverse f^-1
b).graph f(x) and its inverse

c).restric the domain of f so that f^-1 is also a function

d). with the domain of f restricted, sketch a graph of f and f^-1

k so what i did is first i did the inverse:

f(x)=x^2+3
so what i did is let

x=y^2+3
x-3=y^2
y=+-sqrt/x-3

Then i graphed the normal equation and it turned our fine, but i need help drawing the mirror image(with table of values and steps on how to do so)

also someone help with part c and d in detial please

Thank You

 

[tkhunny]

Very, very carefully, think about these two events:

1) In your first sentence, you used the word "function".

2) Way down on the bottom, you wrote the symbol "+-"

I think you may have forgotten what you said in the first sentence.

 

[stapel]

1) In your first sentence, you used the word "function".
2) Way down on the bottom, you wrote the symbol "+-"

So the inverse of the original function is not, itself, also a function. That is to be expected when one begins with a quadratic, and is probably why one of the steps in the exercise is to restrict the domain of the original function so as to be able to create an inverse which is a function. (That is, to get rid of the "plus-minus" from taking the square root.)

a) The equation is given as "y = x² - 3", but you worked with "y = x² + 3". Which is correct? (Your steps look fine, though.)

b) The original function is an upward-opening parabola. Does this obey the Horizontal Line Test for invertible functions? The inverse, being two square roots (that is, two graphs), looks like a "sideways" parabola. Does this obey the Vertical Line Test for being a function?

c) This is telling you to split the parabola at its vertex, and pick one half of the graph. By allowing x to be, say, only zero or less, what sort of graph do you get for the original function and for the inverse? What happens to your "plus-minus"?

Hope that helps!

Eliz.