Thread: Graphing a function given instructions (1st/2nd Derivatives and Intervals)

1. Graphing a function given instructions (1st/2nd Derivatives and Intervals)

Hi you guys, I definitely read the rules and I know it says not to post too many questions at once. But I have a final tomorrow and need your help and patience! I know you all could be doing anything else right now but if you can help I'd greatly appreciate any advice.

This problem involves graphing (see below)

olivi.JPG

The part of the problem that I highlighted sort of confused me. On that part I figured out the sign of f''(x) over the interval (-2, 1) but then wrote the opposite on the actual graph. It seemed to make sense when deciding whether that section of the graph should be convex/concave up/down.

I tried following the instructions and came up with this graph of f(x)

Graph 001.jpg

That was my attempt, it is supposed to follow this example of drawing a graph based on the signs of f'(x) and f''(x) over specified intervals.

inflection.JPG

THANK YOU!

2. In evaluating your graph, the derivative is not zero on (1,2). Since it is given as zero, the value at x = 1 must be the same as the value at x = 2, i.e. f(x) = 2 on the interval (1,2). Also since the derivative (where it exists) is negative on (2,4) and positive on (0,1), and the value of the function is less than or equal to 1 in (-2,0), the maximum value of the function is 2 on the interval (-2,4).

You might also want to look at the interval around x equal 3 where the graph would turn from concave to convex at x equal to 3. That is, the second derivative changes from negative to positive.

3. Hi Ishuda! Thank you for taking the time to go over my graph.

Originally Posted by Ishuda
In evaluating your graph, the derivative is not zero on (1,2). Since it is given as zero, the value at x = 1 must be the same as the value at x = 2, i.e. f(x) = 2 on the interval (1,2). Also since the derivative (where it exists) is negative on (2,4) and positive on (0,1), and the value of the function is less than or equal to 1 in (-2,0), the maximum value of the function is 2 on the interval (-2,4).
Wow of course, since the derivative is the slope of the tangent over that interval, it being zero means that the slope is zero. Got it! Or at least I think I do... I took a second stab at it. (apologize for the crudeness)

Graph2 001.jpg

Originally Posted by Ishuda
You might also want to look at the interval around x equal 3 where the graph would turn from concave to convex at x equal to 3. That is, the second derivative changes from negative to positive.
I think I had the right idea there, I didn't draw it very well on the last graph, but here I tried to make the graph look like its was changing from concave to convex at the inflection point at x=3

I hope I did it right this time. I had a question now though. Since the maximum value of the function is 2 on the interval (1, 2) does that mean that entire interval is a global max? I faintly recall something about this situation meaning there's an infinite amount of global maximums, is that right or am I confusing something?

Thank you again for your time!

4. Looks better. Just a couple of comments; you have f'' marked as negative on your sketch in (1,2). Since f' is zero in that interval, so is f''. You have it sketched correctly, just marked incorrectly. Also, your sketch also looks like the right hand derivative is zero at 2 when you have it marked as negative (which is correct) and the same sort of thing at x=4 from the left.

About your question: A maximum (or minimum) is a y value. The place where that takes place is an x value. So, in our case, there is only one global maximum (2) but there are an infinite number of x's where that occurs (all the x's in the closed interval [1,2])

Oh, and probably just a typo but the derivative is the slope of the curve, not the slope of the tangent. The slope of the tangent is the second derivative.

5. Originally Posted by Ishuda
Looks better.
Couldnt have done it without ya!

Originally Posted by Ishuda
Just a couple of comments; you have f'' marked as negative on your sketch in (1,2). Since f' is zero in that interval, so is f''. You have it sketched correctly, just marked incorrectly.
Did not know that, that makes sense now seeing as how there suspiciously are no instructions for f'' on the interval (1,2) in the problem. And also makes sense because of the revelation you helped me to realize below.

Originally Posted by Ishuda
Also, your sketch also looks like the right hand derivative is zero at 2 when you have it marked as negative (which is correct) and the same sort of thing at x=4 from the left.
Not quite clear on what you mean by this . Are we talking about my sketch right at the red dot where f(2) is? Are you saying it should be like an open circle and start decreasing right at that point? And the same for x=4, an open circle?

Originally Posted by Ishuda
About your question: A maximum (or minimum) is a y value. The place where that takes place is an x value. So, in our case, there is only one global maximum (2) but there are an infinite number of x's where that occurs (all the x's in the closed interval [1,2])
Crystal clear, thank you!

Originally Posted by Ishuda
Oh, and probably just a typo but the derivative is the slope of the curve, not the slope of the tangent. The slope of the tangent is the second derivative.
Why didn't anyone just say that?! lol thank you Ishuda, I just realized I wasnt even exactly sure what a second derivative IS. The slope of the tangent, and the tangent is the slope of the curve which is also the derivative.
So if f'(x)= 0 then f''(x) is going to = 0
Thanks a million.

6. Originally Posted by Taliaferro
....
Not quite clear on what you mean by this . Are we talking about my sketch right at the red dot where f(2) is? Are you saying it should be like an open circle and start decreasing right at that point? And the same for x=4, an open circle?

...
First my terminology is probably explained better with an example: Aside from the problem here, suppose we have a function which is given by

f(x) = x + 1 if x>=1
and
f(x) = x2 + 1 if x<=1

Then
f'(x) = 1 if x > 1
and
f'(x) = 2x if x<1
Although, the derivative is not defined at x=1, the value is equal to 1 everywhere to the right of one. So at x=1 we call the right hand derivative 1. One the left hand side the derivative approaches 2 as we get close to x=1. So at x=1 we say the left hand derivative is 2.

So now back to your sketch: yes, at the read dot at f(2) it looks like the right hand derivative is zero at x=2 and it starts decreasing at that point. Instead, the slope should be negative and it should already be decreasing at that point. The same sort of situation exists at x=4 except it is the left hand derivative.

BTW: We can also note that if a derivative does exist at some x value, then the right hand derivative must equal the left hand derivative.