Question

What is the range of f(x)=√(4-3x^2)
and how do you find it?

Answer 1

Can a square root be a negative number?

Answer 2

yeah because i mean the square root of 4 is 2 and - 2 right?

Answer 3

hm.

Typically you say that the square root of x is positive... the range is zero and positive real numbers.

But I have to admit I am temporarily confused, since you are right, (-2)^2 = 4

Answer 4

well i feel like that can not have what is being squared be a negative because you can not take the square of a negative because that doesn't exist.

Answer 5

The deal is that the domain for y = sqrt(x) is real numbers greater than or equal to zero.

You can't put in a negative number as x and keep y as real numbers...

So given that x must be real and >= 0, the range for y is also real numbers >= 0

Answer 6

The range is the allowable results generated from the allowable inputs... the domain.

So since x = -2 is not in the domain, you don't have to consider it... the x's must be >= 0, so the only resulting y's are also real and >=0

Answer 7

back to your real problem...

you should first find the domain by finding the x values that make the expression under the square root "legal"... that expression must be >=0.

Answer 8

its not a problem of x being negative or not because a negative number squared is always positive. I think x has to be small enough so that you dont take the square of a negative number.

Answer 9

yes :) in your real problem, that is the situation... I was sidetracked on the sqrt(4) issue.

x must be small enough to avoid the negative under the root... and notice, x itself can be negative, as long as x^2 is still small enough.

Answer 10

the domain (-∞, 4/3)

Answer 11

there is a nice alternative for finding range, here

Answer 12

I don't think you can choose really large negative x's... makes the square root argument negative.

maybe it's just between (-4/3, 4/3) ?

Answer 13

@ganeshie8 was there a link? I'm totally interested... there has to be a better way than what I am doing!!

Answer 14

@ganeshie8 i second that

Answer 15

im not sure of links.. mukushla taught me this few days back :

lets say, \(\sqrt{4-3x^2} = k\)

square both sides,

\(4-3x^2 = k^2\)

\(-3x^2+(4-k^2) = 0 \)

Answer 16

since we are looking for real x, \(b^2-4ac >= 0 \)
=>

\(0-4(-3)(4-k^2) >= 0 \)

\(-k^2 >= 4 \)
\(k^2 <= 4\)
\(k <= 2\)

Answer 17

since \(k\) cannot be negative, the range is [0, 2]

Answer 18

interesting, it is...

Answer 19

but how can it be 2 if that would be taking the swuare root of negative 12?

Answer 20

yea it helped me on many occasions with my problems of finding range

Answer 21

[0,2] is the range... it's the range of y values, not the part under the sqrt

Answer 22

oh my gosh.... yeah you are so right i was totally comfised for a second i was thinking domain!!!