Intermediate Value Theorem Excercise

[shelovesmath]

If f(x) = x^2 + 10sinx, show that there is a number c such that f(c) = 1000

Well since I was not given an interval to find a root on, I decided to set the equation equal to 1000 and solve it.

My work:
f(x)=x^2 + 10sinx=1000
x^2 = 1000-10sinx
x= +/- (1000 - sinx)^1/2

 

[nicksauce]

That is a transcendental equation... you're not going to be able to find an analytic solution. My suggestion: Since 10sin(x) is bounded between -10 and +10, you know that the answer will be in the interval √990 and √1010. Now apply the intermediate value theorem to that interval.