y=x√(1-x²)
Domain?
y=x(1-x²)^(1/2)
y=x/[(1-x²)^(-2)]?
x≠±1
Symmetry?
f(3)=192
f(-3)=-300
-f(-3)=300
no symmetry?
For your domain you may only have
-1 < x < +1
Suppose x = 5
then y = 5/√-24 which is undefined.
You only considered when the denominator is zero.
BTW, how did you get those f(x) values,
e.g. f(3) = 192
I get f(3) = 3/√-8
Then isn't the domain -1≤x≤1, since when I plug in -1 and 1, the answer is 0.
For symmetry, I plugged 3 into y=x/[(1-x²)^(-2)], which I'm guessing is wrong. Since domain is -1≤x≤1, I switched the number to 0.5, and found that it is symmetrical about the origin.
Thank you!
Yes, you are right, I should have used the
≤ symbol.
And yes, there is symmetry about the origin.
To determine the domain of the function f(x) = x√(1-x²), we need to consider the restrictions on the values of x that will result in a real output.
In this case, the expression under the square root, (1-x²), should be greater than or equal to zero since taking the square root of a negative number is undefined in the real number system.
So, we solve the inequality 1-x² ≥ 0:
1 - x² ≥ 0
To solve this inequality, we can factor it as (x+1)(x-1) ≥ 0. The critical points are x = -1 and x = 1.
We can now construct a sign chart to determine the intervals where the inequality holds true:
x | (x+1)(x-1)
------------------------
x < -1 | - -
-1 < x < 1 | + -
x > 1 | + +
Based on the sign chart, we can see that the inequality holds true for intervals x < -1 and x > 1. Therefore, the domain of the function is x ≠ ±1.
Moving on to the symmetry of the function, we can check if it exhibits any symmetry by observing the outputs for different inputs.
Given that f(3) = 192 and f(-3) = -300, we can see that the function is not symmetric with respect to the y-axis (f(y) ≠ -f(y)).
Therefore, the function does not exhibit any symmetry.