For the equation x+sin(x)=1 does the intermediate value theorem show at least one solution on the interval [0,pi/6]?

Question

For the equation x+sin(x)=1 does the intermediate value theorem show at least one solution on the interval [0,pi/6]?

a) Yes it shows there must be at least one solution

b) No it is not conclusive

c) No it shows no solutions

Answer 1

f(0) =0 and f(pi/6) = .5+ .52= 1.02

since the function is continuous then it must cross y=1 at least once so the answer is choice A

Answer 2

At 0, the value of the function is 0+ sin(0) = 0+ 0 = 0.

At π/6, the value of the function is π/6 + sin(π/6) = π/6 + 1/2 > 1/2 +1/2 = 1.

Since the function is continuous on [0, π/6] , by the Intermediate Value Theorem it must pass through any y-value between f(0) and f(π/6) or between 0 and π/6 + 1/2. Since this second number is greater than 1, there must be a point on that interval where f(x) = 1, so the answer is a.