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### Course: Calculus, all content (2017 edition) > Unit 3

Lesson 8: Sketching graphs using calculus# Curve sketching with calculus: polynomial

Sal sketches a graph of f(x)=3x⁴-4x³+2 including extremum and inflection points. Created by Sal Khan.

## Want to join the conversation?

- How can the slope be zero if the graph is concave upwards? My brain hurts :((30 votes)
- Concave up means that the slope is increasing. If the slope was below zero and now it's zero, it has increased because 0 is a higher number than a negative number.(55 votes)

- what is the difference between a transition point and an inflection point?? I seem to have missed out on some of the things that Sal mentioned in the earlier lectures... Please help!(20 votes)
- There is no difference in this case. An inflection point (or point of inflection) is the point at which the concavity of the graph changes sign. In this case, the second derivative test is inconclusive, meaning that we must use a difference scheme to determine if x = 0 is in fact an inflection point.(29 votes)

- @5:49In previous videos , Sal analyzed the concavity of a function that he substituted the inflection points in second derivative and check the sign. why did he check the second derivative by using critical points not inflection points?(8 votes)
- He was using the second derivative test to check if those 2 critical points were relative minimum or maximum values on the graph. If the first derivative is equal to 0 and the second derivative is greater than 0 we know it's a relative minimum value, if the second derivative is less than 0 we know it's a relative maximum value, and if the second derivative equals 0 it's inconclusive.(4 votes)

- Why did Sal need to take the second derivative to find the inflection points? Couldnt he have used the first derivative? Im a bit confused.(7 votes)
- Inflection points are where the first derivative has relative max/mins (where the slope of the tangent line of the first derivative =0). He could have used the first derivative but not easily if he did it analytically. You can find points of inflection by looking at the graph of the first derivative, or by solving the 2nd derivative. (At least as far as I know...)(5 votes)

- how does a function look like (in graph) when the slope of a function is 0 and acceleration is 0?

I just cannot visualize it. ;o;(2 votes)- A function that 0 slope and 0 concavity (acceleration) is just a horizontal line.(12 votes)

- Instead of the lengthy discussion starting at9:00, could the third derivative have been used at 2/3 to determine that the second derivative changes sign in the way discussed?(3 votes)
- I know this is a year late but, for future viewers, yes you could have. If the third derivative at 2/3 is not zero, you would know that it's an inflection point. Furthermore, if f'''(2/3) was positive (which it was) you would know that the slope is concave downwards below that point and concave upwards above that point. (You would know that the slope of the second derivative at that point is positive)(7 votes)

- How can I evaluate the derivative of a function at an indicated point?

Please help!(3 votes)- You just take the derivative of that function and plug the x coordinate of the given point into the derivative.

So say we have f(x) = x^2 and we want to evaluate the derivative at point (2, 4). We take the derivative of f(x) to obtain f'(x) = 2x.

Afterwards, we just plug the x coordinate of (2,4) into f'(x). So, what we get is basically f'(x) = 2*2 which equals 4, and that's the slope of the original f(x) at the point (2, 4). The y coordinate, if they give you one, is just extraneous information, it's not needed in this problem.(6 votes)

- how do you get the x and y intercept?(2 votes)
- To find the y-intercept, you make all x-values equal to 0 and solve for y. Inversely, to fine the x-intercept, you make all y-values equal to 0 and solve for x.(6 votes)

- Guys I need some help with this: if you graph ((x+1)*lnx)/(x-1), the graph shows that when x=1, y=2, but why is it not undefined? This is so confusing for me.(2 votes)
- That's because the discontinuity is a "point discontinuity" or a "removable discontinuity". Graphical programs usually don't display this kind of discontinuities.(5 votes)

- I am having a hard time letting everything sink in. My techer used the first derivative test, but you used the second derivative test to find the concavity on a point, the increasing & decreasing intervals, and the inflection points. And are all the critical points either a minimum, maximum or a point of inflectin; or can they have other properties or none at all.

When we plug in the critical points(from the 1st derivative) into the equation of the second derivative are we expecting to identify the concavity, or is there any other proprties we are seeking(other than undetermined points)? And how do we decide which number to plug in, the 2nd dervative, to find the point of inflections; do we use the critical points of the 2nd dervative, or do we add all the undetermined points we got from the critical points of the 1st derivative?

PLEASE, HELP!😪😪(2 votes)- We want to find the maxima, minima, and points of inflection. Maxima and minima can only occur when the first derivative is 0 or undefined (like at x=0 for y=|x| ). Points of inflection only occur when the second derivative is zero.

So all maxima and minima occur when the derivative is zero or undefined. We call these critical points. But not every instance of f'=0 or undefined is a maximum or minimum (consider x=0 at y=x^3). So if we're given a function, we take the derivative and find all the critical points. This set of critical points includes all of the maxima and minima, but it may also include some other things. So we check them each manually.

To check a critical point manually, we consider the second derivative at that point. If the second derivative is nonzero, we know whether the function is concave up or down, and we must have a maximum or minimum.

If the second derivative is zero, the function is not concave up or down at that point. So we check some nearby points to see whether the concavity changes there. If the concavity changes between negative and positive, we have a point of inflection.(4 votes)

## Video transcript

Let's see if we can use
everything we know about differentiation and
concativity, and maximum and minimum points, and inflection
points, to actually graph a function without using a
graphing calculator. So let's say our function,
let's say that f of x is equal to 3x to the fourth minus
4x to the third plus 2. And of course, you could always
graph a function just by trying out a bunch of points, but we
want to really focus on the points that are interesting to
us, and then just to get the general shape of the function,
especially we want to focus on the things that we can take out
from this function using our calculus toolkit, or our
derivative toolkit. So the first thing we probably
want to do, is figure out the critical points. We want to figure out, I'll
write here, critical points. And just as a refresher of what
critical points means, it's the points where the derivative
of f of x is 0. So critical points are f prime
of x is either equal to 0, or it's undefined. This function looks
differentiable everywhere, so the critical points that we
worried about are probably, well, I can tell you, they're
definitely just the points where f prime of x are
going to be equal to 0. This derivative, f prime of x,
is going to actually be defined over the entire domain. So let's actually write down
the derivative right now. So the derivative of this,
f prime of x, this is pretty straightforward. The derivative of 3x to the
fourth, 4 times 3 is 12, 12x to the, we'll just
decrement the 4 by 1, 3. Right? You just multiply times the
exponent, and then decrease the new exponent by one, minus 3
times 4 is 12, times x to the 1 less than 3 is 2. And then the derivative of
a constant, the slope of a constant, you could
almost imagine, is zero. It's not changing. A constant, by definition,
isn't changing. So that's f prime of x. So let's figure out
the critical points. The critical points are where
this thing is either going to be equal to 0, or
it's undefined. Now, I can look over the entire
domain of real numbers, and this thing is defined
pretty much anywhere. I could put any number here,
and it's not going to blow up. It's going to give me an answer
to what the function is. So that it's defined
everywhere, so let's just figure out where
it's equal to 0. So f prime of x is equal to 0. So let's solve, which x is,
let's solve-- I don't have to rewrite that, I
just wrote that. Let's solve for when
this is equal to 0. And I'll do it in
the same color. So 12x to the third minus
12x squared is equal to 0. And so let's what we
can do to solve this. We could factor out a 12x. So if we factor out a 12x, then
this term becomes just x, and then-- actually, let's
factor out a 12x squared. We factor out a 12x squared. If we divide both of these by
12x squared, this term just becomes an x, and then minus
12x squared divided by 12x squared is just 1,
is equal to 0. I just rewrote this
top thing like this. You could go the other way. If I distributed this 12x
squared times this entire quantity, you would get my
derivative right there. So the reason why did that is
because, to solve for 0, or if I want all of the x's that make
this equation equal to 0, I now have written it in a form
where I'm multiplying one thing by another thing. And in order for this to be
0, one or both of these things must be equal to 0. So 12x squared are equal to 0,
which means that x is equal to 0 will make this
quantity equals 0. And the other thing that would
make this quantity 0 is if x minus 1 is equal to 0. So x minus 1 is equal to
0 when x is equal to 1. So these are 2 critical points. Our 2 critical points are x is
equal to 0 and x is equal to 1. And remember, those are just
the points where our first derivative is equal to 0. Where the slope is 0. They might be maximum points,
they might be minimum points, they might be inflection
points, we don't know. They might be, you know, if
this was a constant function, they could just be anything. So we really can't say a lot
about them just yet, but they are points of interest. I guess that's all we can say. That they are definitely
points of interest. But let's keep going, and
let's try to understand the concativity, and maybe we can
get a better sense of this graph. So let's figure out the
second derivative. I'll do that in
this orange color. So the second derivative of my
function f, let's see, 3 times 12 is 36x squared minus 24x. So let's see. Well, there's a couple
of things we can do. Now that we know the second
derivative, we can answer the question, is my graph concave
upwards or downwards at either of these points? So let's figure out what,
at either of these critical points. And it'll all fit together. Remember, if it's concave
upwards, then we're kind of in a U shape. If it's concave downwards,
then we're in a kind of upside down U shape. So f prime prime, our second
derivative, at x is equal to 0, is equal to what? It's equal to 36 0 squared
minus 24 times 0. So that's just 0. So f prime prime is
just equal to 0. We're neither concave upwards
nor concave downwards here. It might be a transition point. It may not. If it is a transition point,
then we're dealing with an inflection point. We're not sure yet. Now let's see what f prime
prime, our second derivative, evaluated at 1 is. So that's 36 times 1, let me
write it down, that's equal to 36 times 1 squared, which
is 36, minus 24 times 1. So it's 36 minus 24,
so it's equal to 12. So this is positive, our second
derivative is positive here. It's equal to 12, which
means our first derivative, our slope is increasing. The rate of change of our
slope is positive here. So at this point right here,
we are concave upwards. Which tells me that this
is probably a minimum point, right? The slope is 0 here, but we are
concave upwards at that point. So that's interesting. So let's see if there
any other potential inflection points here. We already know that this is a
potential inflection point. Let me circle it in red. It's a potential
inflection point. We don't know whether
our function actually transitions at that point. We'll have to experiment
a little bit to see if that's really the case. But let's see if there any
other inflection points, or potential inflection points. So let's see if this
equals 0 anywhere else. So 36 x squared minus
24 x is equal to 0. Let's solve for x. Let us factor out, well,
we can factor out 12x. 12x times 3x, right, 3x
times 12x is 36x squared, minus 2, is equal to 0. So these two are
equivalent expressions. If you multiply this out,
you'll get this thing up here. So this thing is going to be
equal to 0, either if 12 x is equal to 0, so 12 x is equal
to zero, that gives us x is equal to zero. So at x equals 0,
this thing equals 0. So the second derivative is 0
there, and we already knew that, because we tested
that number out. Or this thing, if this
expression was 0, then the entire second derivative
would also be zero. So let's write that. So 3x minus 2 is equal to 0, 3x
is equal to 2, just adding 2 to both sides, 3x is equal to 2/3. So this is another interesting
point that we haven't really hit upon before that might
be an inflection point. The reason why is it
might be, is because the second derivative is
definitely 0 here. You put 2/3 here,
you're going to get 0. So what we have to do, is see
whether the second derivative is positive or negative
on either side of 2/3. We already have a
sense of that. I mean, we could try out
a couple of numbers. We know that, you know, if we
say that x is greater than 2/3. Let me scroll down a
little bit, just so we have some space. So let's see what happens
when x is greater than 2/3, what is f prime prime? What is the second derivative? So let's try out a value
that's pretty close, just to get a sense of things. So let me rewrite it. f prime
prime of x is equal to, let me write like this. I mean, I could write like
that, but this might be easier to deal with. It's equal to 12x
times 3x minus 2. So if x is greater than 2/3,
this term right here is going to be positive. That's definitely, any
positive number times 12 is going to be positive. But what about this
term, right here? 3 times 2/3 minus 2
is exactly 0, right? That's 2 minus 2. But anything larger than that,
3 times, you know, if I had 2.1/3, this is going to
be a positive quantity. Any value of x greater than
2/3 will make this thing right here positive. Right? This thing is also
going to be positive. So that means that when x
is greater than 2/3, that tells us that the second
derivative is positive. It is greater than 0. So in our domain, as long
as x is larger than 2/3, we are concave upwards. And we saw that here,
at x is equal to 1. We were concave upwards. But what about x
being less than 2/3? So when x is less than 2/3,
let me write it, let me scroll down a little bit. When x is less than
2/3, what's going on? I'll rewrite it. f prime prime of x,
second derivative, 12x times 3x minus 2. Well, if we go really far left,
we're going to get a negative number here, and this might
be negative. But if we just go right
below 2/3, when we're still in the positive domain. So if this was like 1.9/3,
which is a mix of a decimal and a fraction, or even 1/3, this
thing is still going to be positive. Right below 2/3, this thing is
still going to be positive. We're going to be multiplying
12 by a positive number. But what's going on right here? At 2/3, we're exactly 0. But as you go to anything
less than 2/3, 3 times 1/3 is only 1. 1 minus 2, you're going
to get negative numbers. So when x is less than 2/3,
this thing right here is going to be negative. So the second derivative, if x
is less than 2/3, the second derivative, right to the left,
right when you go less than 2/3, the seconds derivative
of x is less than 0. Now the fact that we have this
transition, from when we're less than 2/3, we have a
negative second derivative, and when we're greater than 2/3, we
have a positive second derivative, that tells us that
this, indeed, is an inflection point. That x is equal to 2/3 thirds
is definitely an inflection point for our original
function up here. Now, we have one more candidate
inflection point, and then we're ready to graph. Then, you know, once you do all
the inflection points and the max and the minimum, you are
ready to graph the function. So let's see if x is equal to
0 is an inflection point. We know that the second
derivative is 0 at 0. But what happens above and
below the second derivative? So let me do our
little test here. So when x is, let me draw a
line so we don't get confused with all of the stuff
that I wrote here. So when x is greater than
0, what's happening in the second derivative? Remember, the second
derivative was equal to 12x times 3x minus 2. I like writing it this way,
because you've kind of decomposed it into two linear
expressions, and you could see whether each of them are
positive or negative. So if x is greater than 0, this
thing right here is definitely going to be positive, and then
this thing right here, right when you go right above x is
greater than 0, so we have to make sure to be very close
to this number, right? So this number,
let's say it's 0.1. You're right above 0. So this isn't going to be true
for all of x greater than 0. We just want to test exactly
what happens, right when we go right above 0. So this is 0.1. You would have 0.3, 0.3
minus 2, that would be a negative number, right? So right as x goes right
above 0, this thing right here is negative. So at x is greater than 0,
you will have your second derivative is going
to be less than 0. You're concave downwards. Which makes sense, because at
some point, we're going to be hitting a transition. Remember, we were concave
downwards before we got to 2/3, right? So this is consistent. From 0 to 2/3, we are concave
downwards, and then at 2/3, we become concave upwards. Now let's see what happens when
x is right less than, when x is just barely, just
barely less than 0. So once again, f prime, the
second derivitive of x is equal to 12x times 3x minus 2. Well, right. If x was minus 0.1 or 0.0001,
no matter what, this thing is going to be negative, this
expression right here is going to be negative, the 12x, right,
you just have some negative value here, times 12, is
going to be negative. And then what's
this going to be? Well, 3 times minus 0.1 is
going to be minus 0.3, minus 2 is minus 2.3. You're definitely going
to have a negative. This value right here is going
to be negative, and then when you subtract from a negative,
it's definitely going to be negative. So that is also going
to be negative. But if you multiply a negative
times a negative, you're going to get a positive. So actually, right below x
is less than 0, the second derivative is positive. Now, this all might have been a
little bit confusing, but we should now have the payoff. We now have the payoff. We have all of the
interesting things going on. We know that at x is equal to
1, we know that at x is equal to 1, let me write
it over here. We've figured out at x is
equal to 1, the slope is 0. So f prime prime is, sorry,
let me write this way. I should have said, we
know that the slope is 0. Slope is equal to 0. And we figured that out because
the first derivative was 0. This was a critical point. And we know that we're dealing
with, the function is concave upwards at this point. And that tells us that this is
going to be a minimum point. And we should actually get
the coordinates so we can actually graph it. That was the whole
point of this video. So f of 1 is equal to what? f of 1, let's go back to our
original function, is 3 times 1, right, 1 to the fourth is
just 1, 3 times 1 minus 4 plus 2, right? So it's 3 times 1 minus 4 times
1, which is minus 1, plus 2, well, that's just a positive 1. So f of 1 is one. And then we know at x is equal
to 0, we also figured out that the slope is equal to 0. But we figured out that this
was an inflection point, right? The concativity switches
before and after. So this is an inflection point. And we are concave below
0, so when x is less than 0, we are upwards. Our second derivative
is positive. And when x is it greater
than 0, we are downwards. We're concave downwards. Right above, not for all
of the [? domain ?] x and 0, just right
above 0, downwards. And then what is f of 0,
just so we know, because we want to graph that point? f of 0. See, f of 0, this is easy. 3 times 0 minus 4 times 0
plus 2, that's just 2. f of 0 is 2. And then finally we got the
point, x is equal to 2/3. Let me do that in
another color. We had the point x
is equal to 2/3. We figured out that this
was an inflection point. The slope definitely isn't 0
there, because it wasn't one of the critical points. And we know that
we are downwards. We know that when x is less
than 2/3, or right less than 2/3, we are concave downwards. And when x is greater than 2/3,
we saw it up here, when x was greater than 2/3, right up
here, we were concave upwards. The second derivative
was positive. We were upwards. Now we could actually figure
out, what's f of 2/3? That's actually a little
bit complicated. We don't even have to
figure that out, I don't think, to graph. I think we could do a pretty
good job of graphing it just with what we know right now. So that's our take aways. Let me do a rough graph. Let's see. So let me do my axes,
just like that. So we're going to want to
graph the point 0, 2. So let's say that
the point, 0, 2. So this is x is equal to
0, and we go up, 1, 2. So this is the point 0, 2. Maybe I'll do it in that
color, the color I was using, so that's this color. So that's that
point right there. Then we have the point x,
we have f of 1, which is the point 1, 1, right? So this point right here. So that's the point 1, 1. This was the point 0, 2. And then we have x is equal
to 2/3, which is our inflection point. So when x is 2/3, we
don't know exactly what number f of 2/3 is. Maybe here someplace. Let's say f of 2/3
is right there. So that's the point, 2/3, and
then wherever f of 2/3 is. It looks like it's going
to be 1 point something. f of 2/3. You could calculate it, if
you like, you just have to substitute back
in the function. But we're ready to
graph this thing. So we know that at x is
equal to 1, the slope is 0. We know that the slope is 0. It's flat here. We know it's concave upwards. So we're dealing, it looks
like this, looks like that over that interval. We're concave upwards. And we know we're concave
upwards from x is equal to 2/3 and on, right? Let me do it in that color. We knew x equals 2/3 and
on, we're concave upwards. And that's why I was able
to draw this U-shape. Now we know that when x is less
than 2/3 and greater than 0, we're concave downwards. So the graph would look
something like this, over this interval. We'll be concave downwards. Let me draw it nicely. Over this interval, the
slope is decreasing. And you could see it, if you
keep drawing tangent lines. It's flattish there, it gets
negative, more negative, more negative, until the inflection
point, and then it starts increasing again, because we
go back to concave upwards. And then finally, the last
interval is below 0, and we know below 0, when x is less
than 0, we're concave upwards. So the graph looks like this. The graph looks like that. And we also know that x is
equal to 0 was a critical point, the slope of 0. So the graph is actually
flat right there, too. So this is an inflection point
where the slope was also 0. So this is our final graph. We're done. After all that work, we were
able to use our calculus skills, and our knowledge of
inflection points, and concativity, and transitions
in concativity, to actually graph this fairly
hairy-looking graph. But this should be kind of what
it looks like, if you graph it on your calculator.