calculus
posted by james on .
In the following problem, suppose f(x) is continuous (and differentiable) function on the interval (0,1). Suppose also that for 0 < x<(1/4) f(x) is negative, for (1/4) <x<1 f(x) is positive, f(1/4)=0, f (2/3)=1, f ' (1/3)= 1, f ' (2/3) =3.
a. If the function G(x) is given by
G(x)= integral from 0 to x f(t) dt
what can you conclude about the maximum or minimum values of G on (0,1)?
b. What is the slope of the tangent line to G(x) at x=(2/3)?
>>>>please show steps<<<<

b. They tell you that the slope (f') there is 3.