Problem:
For each real number x, let f(x) be the minimum of the numbers 4x+1, x+2, and β2x+4. Then the maximum value of f(x) is
Answer Choices:
A. 31β
B. 21β
C. 32β
D. 25β
E. 38β
Solution:
In the adjoining figure, the graphs of y=4x+1, y=x+2 and y=β2x+4 are drawn. The solid line represents the graph of the function f. Its maximum occurs at the intersection of the lines y=x+2 and y=β2x+4.
Thus x=32β and f(32β)=38β.