Find the three zeros for the following function on the interval -5 </= x </= 5
(1 + 50sin(x)) / (x^2 + 3)
when is sin x = -.02 ??? (x in radians I assume)
x = -.02
x = pi + .02 = 3.16
x = -pi +.02 = -3.12
Thank you Damon. The first time I posted this question, someone gave me the answer 3.1216 for the positive zero. But when I double-checked the answer, it didn't work correctly. I figured something was up and wanted to clarify the answer. 3.161593987 works beautifully in solving the problem, thanks again!
You are welcome.
To find the zeros (also known as roots) of the given function on the interval -5 ≤ x ≤ 5, we need to solve the equation when the function is equal to zero.
First, let's set up the equation:
(1 + 50sin(x)) / (x^2 + 3) = 0
To eliminate the fraction, we can multiply both sides of the equation by (x^2 + 3):
(1 + 50sin(x)) = 0
Now, let's solve for the angle x that satisfies this equation.
1 + 50sin(x) = 0
Subtracting 1 from both sides, we have:
50sin(x) = -1
Dividing both sides by 50, we get:
sin(x) = -1/50
To find the angles that satisfy this equation, we need to use inverse trigonometric functions. In this case, we will use arcsin.
arcsin(-1/50) ≈ -0.02 radians, -3.12 radians
The value -0.02 radians is within the given interval -5 ≤ x ≤ 5, while -3.12 radians is not. Therefore, we only have one root within the given interval.
Therefore, the three zeros of the function on the interval -5 ≤ x ≤ 5 are approximately:
1) x ≈ -0.02 radians
2) x ≈ -0.02 radians
3) x ≈ -0.02 radians