Calc 1, f'(x) > 0 if x < 0 ? What does this mean?

[LearninDaMath]

My math calc 1 class starts just 10 minutes after my physics class ends and it's on the opposite side of campus and yesterday I had my first exam in physics and ended up missing the first 3 minutes of my math class lecture. I copied the notes from the board, but didn't hear what he was saying about the graph. He's an excellent professor, but without hearing what he was saying, I can't figure out what the notes are talking about.

Can anyone look at this graph and the formula and decipher what concept/lesson the professor was describing?

perhaps f(0) = -2 doesn't correspond to anything on this graph..it could be from somthing else he had already erased before I got to class, i just don't know.

So I'm trying to figure out what it all means.

 

[SammyS]

My math calc 1 class starts just 10 minutes after my physics class ends and it's on the opposite side of campus and yesterday I had my first exam in physics and ended up missing the first 3 minutes of my math class lecture. I copied the notes from the board, but didn't hear what he was saying about the graph. He's an excellent professor, but without hearing what he was saying, I can't figure out what the notes are talking about.

Can anyone look at this graph and the formula and decipher what concept/lesson the professor was describing?

The title of this thread, f'(x) > 0 if x < 0 ? What does this mean? means that when x is negative, the slope f(x) (well, actually the slope of the line tangent to f(x)) is positive, as in the lower of the two graphs.

 

[LearninDaMath]

The graph shows what looks like two parabolas, both with vertex at (0, 2), one opening upward, the other opening downward. The rest of the information is related to the lower parabola. It says, for example, that if x< 0, then f'(x)> 0. The derivative, f'(x), can be interpreted as "the slope of the tangent line". f'(x)> 0 means all tangent lines have positive slope- are going up to the right. Also "if x> 0, then f'(x)< 90" so for x positive, the derivative is negative which means tangent lines are going down to the right. The short lines on each dot on the graph represent those tangent lines.

HallsofIvy, thanks

I quoted you to this thread because I asked for deletion of the other thread since I accidently posted it in the wrong forum section and didn't want to lose your response upon deletion. I'm reading your response now.

 

[tiny-tim]

Hi LearninDaMath!

(I assume you're ok with the main part …

demonstrating how the sign of f' works, with two examples, one with f'' > 0 and one with f'' < 0 ?)

I'll guess that that should read f''(0) = -2

 

[LearninDaMath]

I think I understand.

So if I have any function, say, f(x) = 2x^2 + 5x

and I take the derivative: f'(x) = 4x + 5

then for any value of x I choose for the independent variable: say, x = -7,

then f'(x) = 4(-7) + 5 = -23

and so f'(x) < 0 so the slope of the tangent is negative at x = -7

Is this a correct description of the concept here?

Hi Tiny-Tim,

So the first three functions at the top of the notes should be double primes instead of regular functions?

 

[tiny-tim]

Hi LearninDaMath!

… and so f'(x) < 0 so the slope of the tangent is negative at x = -7

Is this a correct description of the concept here?

Yes, wherever f'(x) < 0, the slope is negative

Hi Tiny-Tim,

So the first three functions at the top of the notes should be double primes instead of regular functions?

No, only the first one.

The other two say that, on that particular graph (the top one), f'(x) is negative on the left, and positive on the right (as you can see from the graph).

 

[LearninDaMath]

ah, moment of clarity, the notes make sense now :) thanks