Curious about a functions question?

Question

If I have a function f(x) = sqrt(400-x^2)

and I strictly require it to be valid for real numbers, how many integers can I obtain for values of x?

You don't have to work this out, I am just curious about the method, because I cannot really think of a way to do this.

Answer 1

How many perfect squares are there from 0 to 400?

Because the function will give you an integer if, and only if, the "argument" of the square root is a perfect square.

The squares are:

0, 1, 4, 9, 16... up to 400

Then find the x, so that 400 - x^2 = a perfect square from that list.

Obviously, you stop the list when you hit any integer that is greater than sqrt(400), since x^2 can never be negative.

Therefore, there are 21 integers you can get:

0, 1, 2, 3, 4, 5... up to 20

f(0) = √(400 - 0^2) = √(400 - 0) = √(400) = 20

f(√39) = √(400 - (√39)^2) = √(400-39) = √(361) = 19

f(√76) = √(400-76) = √(324) = 18

and so on

√39 is a perfectly legitimate "real number", even though it is not a rational number nor an integer.

When using √39 as the input to the function, it spits out an integer.

Answer 2

Well first off

x^2 ≤ 400, so we do not have imaginary numbers.

We can only obtain integers from perfect squares, so numbers of the form

k^2, where k is an integer. The sequence of perfect squares are

0, 1, 2, 4, 9, 16, 25, 36, ....

So, knowing the above

400-x^2 = k^2

Also, note that 20^2 = (-20)^2 = 400.

Answer 3

sq roots are only real when the radicand is Positive

400 - x^2 > or = 0

x ^2 (< or =) 400

-20 (< or =) x (< or =) +20

so x is all integers from -20 to + 20

--- including 0 this will be 41 integers

Answer 4

-20 < x < +20

Any values out side this range will give a negative number to 'square root' Y ou cannot ;square root' negative numbers without going into the Realms of 'IMAGINARY' numbers.

Answer 5

41 integers