Question

calculus I ,

Who can help me solve 15 ? https://www.dropbox.com/s/h5qc8nddfj92i6g/Screenshot%202016-01-06%2000.04.07.png?dl=0

Answer 1

particularly to explain me (c) and also tell me why we need closed interval on (a) if at the exact value we get a slope of 0

Answer 2

looks interesting but opening attachments on the net is a no-no for me

Gyazo is a way around that!!!

Answer 3

How is Gyazo different from dropbox?

Answer 4

good question. mea culpa.

Answer 5

So what's the issue here?
You're wondering why they stipulated `open` interval in (c), but not in (a)?

Answer 6

I think it's because ...
a function can change concavity as a singularity, like an asymptote.
In that case, you wouldn't include that boundary in your interval, right?

I think that's it... something like that...

Answer 7

THE GENERAL THEORY:
\(\color{#000000 }{ \displaystyle y=f(x) }\)

then, for whatever function \(f(x)\), we have:
\(\color{#000000 }{ \displaystyle y'=f'(x) }\)
\(\color{#000000 }{ \displaystyle y''=f''(x) }\)

If the function is increasing, then it must have
a positive slope. This is the same as having the
derivative greater than 0. Thus, any \(\color{#000000 }{ \displaystyle x }\) that will
satisfies, \(\color{#000000 }{ \displaystyle f'(x)>0 }\), is the \(\color{#000000 }{ \displaystyle x }\) at which the function
is increasing.

If the function is decreasing, then it must have
a negative slope. This is the same as having the
derivative smaller than 0. Thus, any \(\color{#000000 }{ \displaystyle x }\) that will
satisfies, \(\color{#000000 }{ \displaystyle f'(x)<0 }\), is the \(\color{#000000 }{ \displaystyle x }\) at which the function
is decreasing.


Concavity is the same!! (I'll explain what I mean).

Just as the 1st derivative \(\color{#000000 }{ \displaystyle f'(x) }\) determines if
\(\color{#000000 }{ \displaystyle f(x) }\) increases/decreases/neutral (i.e. =0),
So does the 2nd derivative \(\color{#000000 }{ \displaystyle f''(x) }\) determine if
the 1st derivative \(\color{#000000 }{ \displaystyle f'(x) }\) (the functions' slope),
increases.

Concave up - The slope of the function is increasing.
(e.g. as regards to position, the motion accelerates)

Concave down - The slope of the function is decreasing.
(e.g. as regards to position, the motion decelerates)

When \(\color{#000000 }{ \displaystyle f''(x)=0 }\), then the function has an inflection point.
Provided, that tbhe concavity actually changes.
(e.g. as regards to position, the motion stops decelarating and starts to accelerate, or vice versa)

(just in case there is any misunderstanding)

Answer 8

OK, so for (c) you will need to have an open interval not closed, because you don't want to include the actual x-solutions at which f''(x)=0.

Also, for (a) you need an open interval, because if f'(x)=0, the function is not increasing.

Answer 9

Did it say that you need closed intervals for (a)?
As I see, it didn't...

Answer 10

Parabola, though, has a constant concavity (that entirely depends on the leading coefficient; If you differentiate any parabola twice you will see what I mean)