f(x) is continuous on [2, 4] and differentiable on (2,4). If f (2)= 6 and f(3) = -2 and f(4) =6 :

Question

a. How many solutions are guaranteed?

b. Does f(x) =2 for a value of x in the interval [2, 4]?

c. Can f(x) = 8 for some value of x in the interval [2, 4]?

Answer 1

a) since there are 2 sign changes, there are at least two places where the function crosses the x-axis; so therefore there are at least 2 solutions

b) yes the MEan value theorem guarantees there is x in the interval so that 2 = f(x) = y

c) nothing can be concluded about the max and min values of the function; all we are told is continuity