If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 5

Lesson 9: Sketching curves of functions and their derivatives

# Curve sketching with calculus: logarithm

Sal sketches a graph of f(x)=ln(x⁴+27) including extremum and inflection points. Created by Sal Khan.

## Want to join the conversation?

• Instead of saying ". . just less/greater than. . ."
Wouldn't it just be "just in that sign direction until the critical value?"
So at , Instead of saying, "So when x is slightly less than zero. . " Would it be the same as saying, "When x is between 0 and -3. . . "?
(If there are no more critical points between the value and +/- infinity, then you would just say, "When x is between 0 and negative infinity. . "_
If not, I would like to know why.
Thanks. • You are correct. I think Sal was just trying to stress the meaning of the inequality.

In general there are many ways to say the same thing. Some ways are more formal, some are more standard, some are easier to understand. A good mathematician is always thinking about the best way to say something just like you are.
• I don't completely understand why x = 0 is a minimum point. Sal's logic does make sense (~) that since f"x is positive at both sides of 0, it is concave up all around and x = 0 is a minimum, but in the last video Sal said that in order to verify that a critical point is a min/max you should take the second derivative at that point. If the second derivative is positive it's a min, if it's negative it's a maximum and if it's zero it is neither but may be a potential inflection point. But, if you take the second derivative of zero, in this case, it is zero, so with that logic shouldn't it not be a minimum? • The local max / min (also called "extrema") occur at the bottom or top of a curve. At a max, the curve stops increasing and changes direction to start decreasing (the slope switches from positive to negative). Thus the slope at that exact point where the change occurs is zero -- this is as high as the curve gets in that local area. For a min, it is the other way around, the curve stops decreasing, turns around and starts increasing (the slope switches from negative to positive). Thus the slope at that exact point is zero.

However, you can have a place where the slope is zero, but the curve does not turn around.

So, anywhere the slope is zero is a possible (but not certain) max or min. You have to check it to make sure. You can check by graphing, or you can check using the second derivative.
• If you found the third derivative of a function and you plug in your inflection point candidates into it and you get 0 for one of them, does that immediately tell you that that specific candidate is NOT an inflection point? • Since we know that from -infinity to -3 and from 3 to infinity, the graph is going to be concave downwards, how did we know that the graph doesn't become steeper and steeper until it's basically a vertical line? This may be a little bit confusing wording, but I thought the graph would take a more "m" shape where it would continue sloping down faster and faster as we approached infinity from the positive/negative directions.

I graphed it and it seems to slow its negative acceleration down until it almost looks like a horizontal line. I'm just a bit confused about how Sal knew this when he graphed it. • If you have a set of possible inflection points, why pick numbers slightly above/below each, instead of any point in between? For example, if you have possible inflections at 0 and 3, why check for positive/negative at both 0 plus a little and 3 minus a little, instead of just checking f''(1) and using it for both? • in the AP calculus AB exam, would there be a question regarding sketching? It seems time consuming.. • This is very confusing.
How did he put everything together?
I thought I had to find critical point with the first derivative and apply the "first derivative test" to find critical points and global/local max/min, and then use the "second derivative test" searching for candidate (new critical points) to be inflection points and to find concavity.
How did he find the min/max just from the 2nd derivative? and how did he manage to reuse the same critical points? • I'm confused as to why when using the product rule [f'(x)g(x) + f(x)g'(x)] to calculate the second derivative. You first define the f(x) to be 4x^3 and g(x) to be (x^4 + 27)^-1. Where then, when calculating the second part of the product rule equation [f(x)g'(x)], does he get the second 4x^3? If f(x)=4x^3 and g'(x)= -1(x^4+27)^-2. Why does he multiply 4x^3 twice and then g'(x). Wouldn't that be f(x)f(x)g'(x)? • I'm so confused.. at about Sal decided to factor out 27·12x^2 - 4x^6 = 0 to 4x^2(27·3-x^4) = 0 .. When I was doing it on my own I multiplied the constants and got 324x^2-4x^6 = 0, factoring out to 4x^2(324-x^4) = 0... Is this an example of where BEDMAS is really important? • Is it possible to figure out if 0 is an inflection point by writting a limit instead of just taking really close to zero numbers? • No. That won't work for an inflection point. There are three requirements for an inflection point.
There is an inflection point at x=c if and only if:
1. f(x) is continuous at x=c
AND
2. f''(c) = 0
AND
3. f''(c-ε) has a different sign from f''(c+ε)
This test always works.

A few (not most) inflection points can be found by the first derivative test:
If f(x) is continuous at x=c ,
AND
If f'(c) = 0
AND
If f'(c-ε) has the same sign as f'(c+ε)
THEN
f(x) has an inflection point at x=c

In other words, if the first derivative is 0 at some point but it is not a max or min, then it is an inflection point. However, most inflection points cannot be found this way because they will be located at some point where the first derivative is not 0.
(1 vote)