How do you find the zeroes of x/(9x^2-4)? I thought the answer would be the sq. root of 4/9, but my book says the answer is zero. Why is this?
To find the zeroes of a function, we need to find the values of x for which the function equals zero. In this case, we have the function f(x) = x/(9x^2 - 4).
To find the zeroes, we need to solve the equation f(x) = 0, which means we need to find the values of x that make the numerator equal to zero.
In this case, we have x = 0 as a possible zero. Let's substitute x = 0 into the function and see if it equals zero:
f(0) = 0/(9(0)^2 - 4) = 0/(0 - 4) = 0/(-4) = 0.
So, x = 0 is indeed a zero of the function.
Now let's consider the other term in the denominator, 9x^2 - 4. We can solve this quadratic equation to find if there are any other possible zeroes.
9x^2 - 4 = 0
Adding 4 to both sides:
9x^2 = 4
Dividing by 9:
x^2 = 4/9
To find the square root of 4/9, we need to consider both the positive and negative square roots because the original equation is quadratic. The positive square root of 4/9 is 2/3, and the negative square root is -2/3.
Therefore, the possible zeroes are x = 0, x = 2/3, and x = -2/3.
However, upon further examination, we find that x = 2/3 and x = -2/3 are not actual zeroes since they make the denominator zero. For these values of x, the function is undefined. So, the only zero of the function is x = 0.
It's important to double-check the domain (values of x for which the function is defined) in cases like these, as it may lead to restrictions on the possible zeroes.