Monday

August 31, 2015
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.

- Calculus (Urgent Help) -
**s_Is_!@#$%^&**, Friday, June 12, 2015 at 3:46pm@S

Who cares? The exam is written on pencil and paper and by the end you can easily determined who cheated. It's their loss. If they are taking calculus in university they will struggle, and that will be their fault. I'm pretty sure you're my teacher, but who cares.