derivative again

Its Like
1. If the first derivative f' is positive (+) , then the function f is increasing (A) .

2. If the first derivative f' is negative (-) , then the function f is decreasing (V) .

3. If the second derivative f'' is positive (+) , then the function f is concave up (n) .

4. If the second derivative f'' is negative (-) , then the function f is concave down (u) .

5. The point x=a determines a relative maximum for function f if f is continuous at x=a , and the first derivative f' is positive (+) for x<a and negative (-) for x>a . The point x=a determines an absolute maximum for function f if it corresponds to the largest y-value in the range of f .

6. The point x=a determines a relative minimum for function f if f is continuous at x=a , and the first derivative f' is negative (-) for x<a and positive (+) for x>a . The point x=a determines an absolute minimum for function f if it corresponds to the smallest y-value in the range of f .

7. The point x=a determines an inflection point for function f if f is continuous at x=a , and the second derivative f'' is negative (-) for x<a and positive (+) for x>a , or if f'' is positive (+) for x<a and negative (-) for x>a .

So depedning on this fact .... you are right ....all answers are possible for f(7)

I believe using the Mean Value Theorem would be best.

a=5, b=7.

Apply the MVT to f' to get
f''(c)=(f'(7)-f'(5))/2 for some c in (5,7).

Then as f''(c)<0 (as c>5), f'(7)<f'(5). f'(5)=20 implies f'(7)<20.

Now apply MVT to f to get
f'(d)=(f(7)-f(5))/2 for some d in (5,7).

Then 2f'(d)=f(7)-20; so f(7)=2f'(d)+20. If we can bound f'(d), say f'(d) is between X and Y, then f(7) is between 2X+20 and 2Y+20. So let's try to bound f'(d).

Apply the MVT to f' on the interval (5,d).
f''(e)=(f'(d)-f'(5))/(d-5) for some e in (5,d). We're given that f''(e)<0 as e>5. Then (f'(d)-f'(5))/(d-5) is <0. Solving for f'(d), we get f'(d)<20 as f'(5)=20 and d-5>0. This seems to me ! the best we can do with f'(d) with the given information. Our "Y" is 20 and there is no X.

Thus, f(7)=2f'(d)+20 gives us that since f'(d)<20, f(7)<2*20+20=60.

If I did everything right, f(7) cannot exceed or equal 60. Hrm, that seems way off your answers though...