Intervals of

Answer the following questions about the function f:[A,I] \to \mathbb R whose graph is shown below.

1. On what intervals is f' < 0?
2. On what intervals is f' > 0?
3. On what intervals is f'' < 0?
4. On what intervals is f'' > 0?
5. At what points is f' = 0?
6. At what points is f'' = 0? (Might be tricky)
7. At what points is f discontinuous and why?
8. At what points is f not differentiable and why?

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1. Intervals of f'<0 =(A,B),(C,E),(G,H)
2. Intervals of f'>0=(B,C),(E,G),(H,I)
3. intervals of f"<0=(B,D),(F,H),(A,B)
4. Interval of f">0=(D,F)
5. Point where f'=0 C,E,G
6. Where does f''=0 on interval (H,I) and points D, F
7. f is discontinuous at point H, because the limit from the left and right do not both correspond with f(h).
8. f is not differentiable at points B, H because neither have a slope so they would not exist in f'(x).

@mearing - That’s a good start! Here are a few comments:

• It looks like you’ve got f'>0 and f'<0 backwards.
• It looks to me like the graph is not quite a straight line on the interval (A,B) so I think that f''<0 there, not equal to zero.
• I added one part on differentiability!

1. f' < 0 on these intervals: (A, B), (C, E), (G, H)
2. f' > 0 on these intervals: (B, C), (E, G), (H, I)
3. f'' < 0 on these intervals: (A, B), (B, D), (F, H)
4. f'' > 0 on these intervals: (D, F)
5. f' = 0 on these points: B, C, E, G, I
6. f'' = 0 on these points: B, C, E, G, I
7. f is discontinuous at the point H because as you approach H from the left and right the limit is not the same.
8. f is not differentiable at the point H because it is not continuous.

@mbradle7 this looks really close! I think you’re a bit off on 6 and 8, though.

On 6, there is a whole interval of points on which $f& has zero second derivative.

I think you’re missing one point on 8.