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# Curve Sketching with Derivatives - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

In order to sketch a graph, you need to know a function's critical points and inflection points. To find the critical points of the function, calculate the first derivative and find its zeros. Then, find the value of the function at those points (by plugging them in to your original function). These points will be your relative maxima and/or minima. Then, to find the inflection points, find the zeros of the second derivative. Remember that these are points at which the function may change from concave up to concave down, and vice versa--this can help you visualize the graph better.

Here’s another curve sketching problem. I’m asked to graph g(x) equals x times the quantity 5 minus x to the 2/3 power. Now I calculated g' and g'' before, and I got 15 minus 5x over 3 times the cube root of 5 minus x and g'' is -60 plus 10x over 9 times (5 minus x) to the 4/3. I need to find critical points and make a sign chart for these derivatives. First let’s find some critical points.

You’ll notice that both derivatives have 5 minus x in them, so both derivatives are going to be undefined when x equals 5. Now, to determine whether x equals 5 is a critical point, I need to look back at my original function to make sure it is defined there. If it’s not defined there, then it doesn’t count as a critical point, the function does not going to have a point to plot there. But x equals 5 is in the domain, this is perfectly defined. So our critical point will be x equals 5.

And what else? Looks like we’re going to have x equals 3, that will make the first derivative zero. So I'll mark that x equals 3 and then over here on the second derivative, notice that x equals 6 will make the second derivative 0. So that’s another important point.

Let’s make our sign chart. First I’m going to mark a row for g'(x) and g''(x). We’re going to draw pluses or minuses to tell me whether the derivative is positive or negative in each part of the interval. We mark these points; 3, 5 and 6. The first derivative is undefined at 5 and it equals 0 at 3. So those are the critical points.

What happens in between? What happens over here? What happens between 5, and what happens to the right of 5? We need to plug in points and see. And so I’ll try x equals 2. I plug in 2 here, I’m going to get 15 minus 10, 5, it’s positive. If I plug in 2 here, I get a positive number. So I get a positive over a positive which will be positive. So g' is positive.

In between 3 and 5 I could try x equals 4. And so I get 15 minus 20, that’s negative. 5 minus 4, positive, to the 1/3 still positive. So negative over positive is negative. And then over here, I could use 6 to test that variable. I have 15 minus 30, that’s negative. 5 minus 6, also negative. I’m going to have a negative numerator, a negative denominator, that’s going to be positive.

And now for g', and you remember that at x equals 5, g' is also undefined. So we need to put a little u here. And g' is zero at x equals 6. Now let me figure out what’s happening in the other intervals.

X equals 3. I’m going to have -60 plus 30, that’s -30. It’s negative on top and then down here, 5 minus 3, 2, positive. So negative over positive it’s going to be negative. It will be negative all the way up until this point. It’s undefined here. We have to look in between 5 and 6 to see what’s happening.

So let’s try 5.5. This is going to be 55 minus 60; it will be negative on top. And 5 minus 5.5 is -0.5 but to the 4/3 poser, it’s going to be positive, because you are going to be cube rooting it and then raising to the 4th power. That’s going to turn our negative into a positive. So it will be negative over positive, negative in here. Let me just put a plus sign up here, because remember the first root is actually positive all to the right of x equals 5. So on forever.

Now let’s take a look at a little bit to the right of 6. Like 7. I’ll get 70 minus 60 which is 10, positive. 5 minus 7, a negative number, but to the 4/3 still positive. So positive over positive is positive. Now let me draw some little tiny curves to give me some idea of what this graph’s going to look like.

It’s going to be increasing because g' is positive, concave down. So something like this. And then decreasing concave down. So like that. Between 5 and 6, it’s increasing concave down and from 6 on, it will be increasing and concave up. So like that. You can see what’s happening here at x equals 5, where these two pieces meet. We’re going to get a little corner or cusp.

In fact if you want to see, if you want to get an idea as to whether it’s a corner or an actual cusp, like a vertical cusp, you can try to think about what happens as x gets close to 5. You see the denominator gets close to zero so the derivative’s going to infinity. It’s going to get vertical there. So this is actually going to come to a vertical point. I’ll have to remember that when I’m plotting points.

We have some special points to plot. We have x equals 3, 5 and 6. And I took the liberty of calculating these before hand. G(3) first of all let’s 3 times 5 minus 3, 2, to the 2/3. 3 times 2 to the 2/3 and I calculated that earlier it’s about 4.8. And I need g(5).

That’s the easy one. You plug 5 and you get 0. So g(5) is 0. G(6), when I plug 6 in, I get -1, but to the 2/3 is going to be +1. So 1 times 6 is 6. Another important point, if you take a look at this you can see that when you plug in 0, you will also get 0. That’s an intercept. So it’s an important point. G(0) is 0.

So I have 4 key points to plot and I'll be drawing my graph.

Let’s start with what happens at zero. It crosses right through (0,0). I’ll draw my y axis in later but this is x equals 0. 3, I’ll put 6 in here and put 5 right next to it. Now remember, to the left of 3, it’s increasing in concave down. So the graph’s going to look something like this.

It’s got to go through the origin, increasing concave down and comes to a critical point here, where it hits a maximum value of 4.8. That value is going to be 4.8. Then it’s going to come down in decreasing concave down to a vertical point here at 5. Then it’s going to increase concave down until it reaches a value of 6. That’s the inflection point, where it changes from concave down to concave up like so. This is the point (6,6), this is the point (3,4.8). You can see that these are x intercepts.

And so this is a pretty good graph of my function. Without a calculator, a pretty good graph of g(x) equals x times 5 minus x to the 2/3.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

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