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RideAnS [48]
3 years ago
11

For a function p(x)= x^2-9, what is the inverse function for the domain [0, infinity\

Mathematics
1 answer:
kkurt [141]3 years ago
7 0
An inverse function is a function that reverses another function, so we have a function called:

p(x)

Then, the inverse function will be as follows:

p^{-1}(x)

Given that y = p(x), we need to isolate x in terms of y:

y =  x^{2} -9
∴ y+9 = x^{2}

So:

x=+\sqrt{y+9} and x=-\sqrt{y+9}

therefore, exchanging variables x and y:

y=+\sqrt{x+9} and y=-\sqrt{x+9}
p^{-1}(x)=+\sqrt{x+9} and p^{-1}(x)=-\sqrt{x+9}

Which are the figures shown below.

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(x - 2)² + (y - 6)² = 4

You can be certain about one thing just by looking at the equation (2, 6) is the center, so, obviously the circle isnt going through this point

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