functions from graphs

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dubb

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Hi, I'm trying to wrap my brain around this one. Can you please help?

Use the graphs of f' and f'' to find critical points, inflection points, intervals on which f is increasing and decreasing, and intervals of concavity, then graph f assuming f(0)=0

(f' is in red, f'' is blue line)
graph.jpg


here's what i see so far: (0,1) f'<0 Decreasing Concave up f''>0, (1,2) f'>0 increasing concave up f''>0.
I'm having trouble understanding the line f'' and how to interpret it. Any help is greatly appreciated. Thanks!
 
dubb said:
Use the graphs of f' and f'' to find critical points, inflection points, intervals on which f is increasing and decreasing, and intervals of concavity, then graph f assuming f(0)=0

(f' is in red, f'' is blue line)
View attachment 4572


here's what i see so far: (0,1) f'<0 Decreasing Concave up f''>0, (1,2) f'>0 increasing concave up f''>0.
I'm having trouble understanding the line f'' and how to interpret it. Any help is greatly appreciated. Thanks!
Click to expand...
These are hard. You're kinda trying to go backwards from the picture. :shock:

Where the function is decreasing, the derivative is negative; where the function is increasing, the derivative is positive; there the function is horizontal (or has a critical point), the derivative is zero.

Working backwards: Where the derivative is negative, the function is decreasing; where the derivative is positive, the function is increasing; where the derivative is zero, the function is horizontal (or there's a critical point).

In your graph, is the derivative ever negative? So is the function ever decreasing? (Note: "Increasing at a slower rate" is not the same as "decreasing"!) Is the derivative ever zero? Where? So where is the function horizontal?

Where the function is concave up, the second derivative is positive; where the function is concave down, the second derivative is negative; where the function has an inflection point, the second derivative is equal to zero.

Working backwards: Where the second derivative is positive, the function is concave up; where the second derivative is negative, the function is concave down; where the second derivative is zero, the function has an inflection point.

In your graph, is the second derivative ever negative? If so, where? So is the function ever concave downward? Same questions for positive and zero.

Doodling, I come up with:

Code: 
f':

increasing, but by less and less,
followed by
increasing, and by more and more:
          
            - - '                     /
       - '                          /
    -'     followed by           -'
  /                           - '
/                       - - '

...and:

Code: 
f'':

concave down,
inflection point, and
concave up:

      - -    -ish,               /
  - '     '                    -
/         followed by  ' - - '

See what you can come up with, based on these bits and pieces. ;-)
 
Looking at the graph, I would say f'' is a straight line or
f'' = a * (x - 1)
since it is zero at x = 1.

Therefore
f' = a * (0.5 * x2 - x + 0.5 * c) = 0.5 * a * ( x2 - 2 x + c) = A (x-1)2
since f' appears to have a double zero at x = 1. In addition, A is positive since f' is non-negative.

Thus, replacing the arbitrary positive constant A/3 by another arbitrary positive constant B, we have
f(x) = B * [ (x - 1)3 + 1]
since f(0) = 0.

One way of interpreting the graphs is to, for example, assume f(x) is a distance traveled by a particle [the exact distance depends on the constant B and a point of reference]. Then f' is a velocity and f'' is an acceleration. So the particle is decelerating for x < 1 (f'' is negative) and accelerating for x > 1 (f'' is positive). It is slowing down to zero speed for x < 1 and speeding up from zero speed for x > 1. The particle is 'behind us' (f is negative) for x < 0 and is 'in front of us' (f is positive) for x > 0.

Edit to add in the constant for f(x), i.e. f(0) = 0.
 
Last edited:
stapel said:
These are hard. You're kinda trying to go backwards from the picture. :shock:

In your graph, is the derivative ever negative? So is the function ever decreasing? (Note: "Increasing at a slower rate" is not the same as "decreasing"!) Is the derivative ever zero? Where? So where is the function horizontal?


In your graph, is the second derivative ever negative? If so, where? So is the function ever concave downward? Same questions for positive and zero.



See what you can come up with, based on these bits and pieces. ;-)
Click to expand...

In your graph, is the derivative ever negative? So is the function ever decreasing? (Note: "Increasing at a slower rate" is not the same as "decreasing"!) Is the derivative ever zero? Where? So where is the function horizontal?

the derivative is negative on interval (0,1) so the function is decreasing, the derivative is is positive on interval (1,2) so function is increasing. the derivative is 0 at x=1, so the function has slope 0.


In your graph, is the second derivative ever negative? If so, where? So is the function ever concave downward? Same questions for positive and zero.

the second derivative kinda throws me off, i'm kinda guessing that it increases with concave down on interval (0,1), from 1,2 it increases concave up. inflection point at 1.
 
dubb said:
the derivative is negative on interval (0,1) so the function is decreasing, the derivative is is positive on interval (1,2) so function is increasing. the derivative is 0 at x=1, so the function has slope 0.
Click to expand...
So it's the blue line that's the first derivative? :wink:
 
stapel said:
So it's the blue line that's the first derivative? :wink:
Click to expand...

Gosh, i'm really trying to wrap my mind around it, the blue line is the second derivative. I'm seeing that as f' decreases/f is negative then f'' is in the negative below 0, and then both f' and f'' increase after x=1.
 
dubb said:
I'm seeing that as f' decreases/f is negative then f'' is in the negative below 0, and then both f' and f'' increase after x=1.
Click to expand...
No. When f decreases, then f' is negative, and so forth:

Where the function is decreasing, the derivative is negative; where the function is increasing, the derivative is positive; there the function is horizontal (or has a critical point), the derivative is zero.

Working backwards: Where the derivative is negative, the function is decreasing; where the derivative is positive, the function is increasing; where the derivative is zero, the function is horizontal (or there's a critical point).

In your graph, is the derivative ever negative? So is the function ever decreasing? (Note: "Increasing at a slower rate" is not the same as "decreasing"!) Is the derivative ever zero? Where? So where is the function horizontal?
Click to expand...
 
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