Posted by **james** on Friday, July 3, 2009 at 11:11pm.

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<<<<

## Answer This Question

## Related Questions

- calculus - Let f be the function defined by f(x)= sqrt(x), 0 <or= x <or= 4...
- Calculus - Show that the function f(x)= x^(3) +3/(x^2) +2 has exactly one zero ...
- Math11 - Hello, I dont know how to do this, please help. Thank you. 1).Does the ...
- calc - Which of the following statements would always be true? I. If f is ...
- Calculus - COnsider g(x)=(8)/(x-6) on (6,13) (a) Is this function continuous on ...
- Calculus - COnsider g(x)=(8)/(x-6) on (6,13) (a) Is this function continuous on ...
- Calculus - COnsider g(x)=(8)/(x-6) on (6,13) (a) Is this function continuous on ...
- Math - a) Is ∫[-1 to 1]e^x^3 dx positive, negative, or zero? Explain. I think ...
- Calculus - Suppose that f(x), f'(x), and f''(x) are continuous for all real ...
- calculus - For what value of the constant c is the function f continuous on the ...

More Related Questions