Problem:
Let f(x)=sinx+2cosx+3tanx, using radian measure for the variable x. In what interval does the smallest positive value of x for which f(x)=0 lie?
Answer Choices:
A. (0,1)
B. (1,2)
C. (2,3)
D. (3,4)
E. (4,5) Solution:
For 0<x<2π all three terms are positive, and f(x) is undefined when x=2π. For 2π<x<43π, the term 3tanx is less than -3 and dominates the other two terms, so f(x)<0 there. For 43π≤x<π,∣cos(x)∣≥∣sin(x)∣ and cosx and tanx are negative, so sinx+2cosx+3tanx<0. Therefore there is no positive solution of f(x)=0 for x<π. Because the range of f includes all values between f(π)=−2<0 and f(45π)=−232+3>−1.5⋅1.5+3>0 on the interval [π,45π], the smallest positive solution of f(x)=0 lies between π and 45π. Because π>3 and 45π<4, the requested interval is (3,4).