Posted by **Mishaka** on Friday, November 11, 2011 at 4:50pm.

Find the three zeros for the following function on the interval -5 </= x </= 5

(1 + 50sin(x)) / (x^2 + 3)

- Calculus -
**Reiny**, Friday, November 11, 2011 at 5:12pm
For the expression to be zero,

1 + 50sinx = 0

sinx = -1/50

x = π +.02 or 2π - .02

x = 3.1616 or 6.263

the period of 50sinx is 2π, so adding or subtracting 2π to the above answers will produce more answers

3.1616+2π ---> beyond domain

3.1616-2π = -3.1216

6.263-2π = -.02

so for -5 ≤ x ≤ 5

x = -3.1216 , -.02, 3.1216

- Calculus -
**Mishaka**, Friday, November 11, 2011 at 5:20pm
Thank you, this helps tremendously!

- Calculus -
**Mishaka**, Friday, November 11, 2011 at 6:57pm
How did you come up with 3.1216 positive. I've tried to come up with this number but I can't figure it out. Also, it doesn't make 1 + 50 sin(x) = 0.

- Calculus -
**Reiny**, Friday, November 11, 2011 at 9:43pm
It was a copy error.

As you can see from the 5th line, I had x = 3.1616

so the final answers are

x = -3.1216 , -.02, and 3.1616

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