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<2Οβ all three terms are positive, and f(x) is undefined when x=2Οβ. For 2Οβ<x<43Οβ, the term 3tanx is less than -3 and dominates the other two terms, so f(x)<0 there. For 43Οββ€x<Ο,β£cos(x)β£β₯β£sin(x)β£ and cosx and tanx are negative, so sinx+2cosx+3tanx<0. Therefore there is no positive solution of f(x)=0 for x<Ο. Because the range of f includes all values between f(Ο)=β2<0 and f(45Οβ)=β23β2β+3>β1.5β 1.5+3>0 on the interval [Ο,45Οβ], the smallest positive solution of f(x)=0 lies between Ο and 45Οβ. Because Ο>3 and 45Οβ<4, the requested interval is (3,4)β.