Wednesday

May 25, 2016
Posted by **Veronica** on Friday, July 25, 2008 at 4:38pm.

1. If ƒ′(x) < 0 when x < c then ƒ(x) is decreasing when x < c.

True

2. The function ƒ(x) = x^3 – 3x + 2 is increasing on the interval -1 < x < 1.

False

3. If ƒ'(c) < 0 then ƒ(x) is decreasing and the graph of ƒ(x) is concave down when x = c.

True

4.A local extreme point of a polynomial function ƒ(x) can only occur when ƒ′(x) = 0.

True

5. If ƒ′(x) > 0 when x < c and ƒ′(x) < 0 when x > c, then ƒ(x) has a maximum value when x = c.

False

6. If ƒ′(x) has a minimum value at x = c, then the graph of ƒ(x) has a point of inflection at x = c.

False

7. If ƒ′(c) > 0 and ƒ″(c), then ƒ(x) is increasing and the graph is concave up when x = c.

True

8. If ƒ′(c) = 0 then ƒ(x) must have a local extreme point at x = c.

False

9. The graph of ƒ(x) has an inflection point at x = c so ƒ′(x) has a maximum or minimum value at x = c.

True

10. ƒ′(x) is increasing when x < c and decreasing when x > c so the graph of ƒ(x) has an inflection point at x = c.

True

So yeah those are the questions (statements), if you can tell which ones are true / false or correct me on what I said that would be helpful. I put what I thought it was so if its wrong, please correct me! Thanks,

Veronica!!

- Calculus (Urgent Help) -
**Damon**, Friday, July 25, 2008 at 5:37pm1. T agree - if x is increasing, f(x) is decreasing

2. F agree - max at x = -sqrt 3, min at x = +sqrt 3

3. False, disagree, it is decreasing but who says it is concave or convex? It could be a straight line with negative slope. - Calculus (Urgent Help) -
**Damon**, Friday, July 25, 2008 at 5:41pm4.A local extreme point of a polynomial function ƒ(x) can only occur when ƒ′(x) = 0.

True AGREE

5. If ƒ′(x) > 0 when x < c and ƒ′(x) < 0 when x > c, then ƒ(x) has a maximum value when x = c.

TRUE I think - DISAGREE looks like /\ - Calculus (Urgent Help) -
**Damon**, Friday, July 25, 2008 at 5:47pm6. If ƒ′(x) has a minimum value at x = c, then the graph of ƒ(x) has a point of inflection at x = c.

False

TRUE - DISAGREE

an extreme value of f' means f" changes sign which means inflection - Calculus (Urgent Help) -
**Damon**, Friday, July 25, 2008 at 5:49pmAgree with 7 8 9 10

- Calculus (Urgent Help) -
**Veronica**, Friday, July 25, 2008 at 5:51pmoh okay, thanks so much! :) i really appreciate the help, especially since u corrected what i did wrong. thanks!

- Calculus (Urgent Help) -
**S**, Monday, May 10, 2010 at 4:17pmI hope you realize by doing this you are helping students taking online calculus cheat on their test, as this is direct content from the test under Unit 6 Activity 9.

- Calculus (Urgent Help) -
**Philip**, Thursday, October 20, 2011 at 1:30amNot only that, but you are wrong on number 6.

A minimum value of f'(x) at x = c does NOT imply an extreme value, although the reverse is certainly true if the sign changes from - to +.

f''(x) could be undefined at f'(c) which would make it a critical value of f'(x) since it is in the domain of f'(x). An example would be if f'(x) = x ^ (1/2).

(0,0) is clearly a minimum value of f'(x), and in fact f(x), but it is not a relative minimum so there would be no inflection point.