Question

AP Calculus AB Help PLEASE! Thanks in advanced...
I need help with these problems
- http://prntscr.com/bd41hs
- http://prntscr.com/bd423g
- http://prntscr.com/bd43o5

Answer 1

@Mehek14 @nincompoop @jigglypuff314 @ganeshie8

Answer 2

@Kainui @jabez177 @Preetha

Answer 3

Allright, AP Calculus Guy, do you agfree we are shown f' (f-prime, the _derivative_ of f), so have to use the derivative to figure out things about the original function f?

Answer 4

Yes

Answer 5

On the first problem. I'm going to ask that you repost the other two problems separately.

Answer 6

ok that's fine

Answer 7

OK, what does it mean when the derivative of something is zero?

Also, what does it mean when the second derivative of something is zero?

Answer 8

there's no derivative?

Answer 9

Whoops, I misunderstood your answer...

Answer 10

So, something not existing, and something being zero are not the same thing.

Derivative is the _slope_ of the original function, right? IN GENERAL, what kind of line has slope m=0?

Answer 11

a straight horizontal line

Answer 12

So at x=2 and x=5 ,the derivative is zero, right. So those parts of _f_ are horizontal, if only briefly. (Technically, the TANGENT is horizontal, right?)

Answer 13

yeah i understand that

Answer 14

So, what is the graph doing to the _left_ of x=2? What is the slope, and what does that mean for the value of _f_ ? (Bigger or smaller than at x=2)?

Answer 15

it's increasing, the slope is positive and not sure what that means for f

Answer 16

Let's try that again. Reading the graph of f', is the _value_ of f' positive or negative at x=1.95? How about x=2.05?

Answer 17

At -1.95 f' is positive and at 2.05 it's negative

Answer 18

I think you are confusing the _values_ of f' with the _slope_ of f' . We need to worry right now about the _values_ of f'

At x=-2, the value of f' is -3. It increases to f'(x=2) = 0, then decreases again to f'(x=4) = -2.5

Remember that when we find positive or negative values of f' , it tells us if the slope of _f_ is negative (moving down-and-right) or positive (moving up-and-right)

So the slope of _x_ is negative, then zero, then negative again around x=2

Answer 19

Do you see that now?