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The unit circle with center at the origin is a relation but not a function.

a. Find the two functions which are semicircles of the unit circle, and determine
their domains and ranges?
b. Are these functions onto functions? Justify.
c. Find the two functions which are lie in one of the semicircle in part a, and
determine their domains and ranges?
d. Are these functions one to one correspondence functions? Justify.​

User Abhilb
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1 Answer

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28 votes

Explanation:

the formula of the "relation" is

x² + y² = r²

this is for circle with its center at the origin.

now that gives us

y² = r² - x²

y = ±sqrt(r² - x²)

a. this ± of the square root gives us the 2 functions.

y = sqrt(r² - x²)

y = - sqrt(r² - x²)

as it is a circle, the ranges and domains are the same.

domain is the allowed number interval for x, and range is the number interval for y (the function results).

both are

-r <= x <= r

outside of that range the square root argument becomes negative, and that is undefined for R (the set of real numbers).

b. if we want to see it that way, we can say that

f(x) = r² - x²

g(x) = sqrt(x)

and so

sqrt(r² - x²) = g(f(x))

a function into function.

c. let's pick the "positive" semicircle

y = g(f(x)) = + sqrt(r² - x²)

f(x) = r² - x²

the domain is all x (-infinity < x < +infinity)

the range is (-infinity < y <= r²), as negative and positive larger x will turn f(x) negative. the functional results cannot get bigger than r² (for x=0).

g(x) = +sqrt(x)

the domain is all positive x incl. 0 (0 <= x < +infinity), as sqrt is not defined for negative values. at least not in R.

the range is also all possible positive values incl. 0

(0 <= y < +infinity).

d.

A one-to-one correspondence function is a function between the elements of the two sets of domain and range, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set.

that means, every x value creates exactly one y value, and no y value occurs more than once.

this is no true for

f(x) = r² - x²

because for -r <= x <= r we get the same y for -x and +x.

it is true for

g(x) = sqrt(x), as no 2 different numbers have the same square root.

User Swigganicks
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