 ###### 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.

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# Critical Points - Problem 3

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.

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Critical points of a function are where the derivative is 0 or undefined. To find critical points of a function, first calculate the derivative. Remember that critical points must be in the domain of the function. So if x is undefined in f(x), it cannot be a critical point, but if x is defined in f(x) but undefined in f'(x), it is a critical point.

On a graph, critical points can mean one of two things: that there is a horizontal tangent at that point (if f'(x)=0 at that x), or there is a vertical tangent at that point (if f'(x) is undefined at that x).

We are talking about finding critical points for a function. Now here’s another example. Consider h(x) equals x to the 2/3 times the quantity x minus 4. And I’ve graphed that function here. Find the critical points of h and explain their geometric significance.

Let’s start by recalling that a critical point, is a point with a derivative equal to 0 or is undefined. So I want to look at the derivative of this function. And this function is a product. So I’ll use the product rule, the product rule on this.

So it's first times the derivative of the second. And the derivative of the second is just 1, plus the second times the derivative of the first. I’ll use the power rule on x to the 2/3. So it’s going to be 2/3x to the 2/3 minus 1. And 2/3 minus 1 is -1/3. So this is 2/3x to the -1/3. Now, let me simplify this a little bit.

I have x to the 2/3 here, plus, now this x to the -1/3 it means 1 over x to the 1/3. I have a 3 in the denominator as well. So I’m going to show you what’s in the denominator here, there’s 3 and x to the 1/3.And in the numerator there’s a 2 and x minus 4. Now, in order to find to critical points, I need to factor this completely. But when you have to also get a single fraction.

I have two different expressions here I need to combine them to a single one. The way to do that is to get a common denominator. So I need to get this guy to have the same denominator as this guy. This denominator is 3x to the 1/3. So I multiply top and bottom of this. Think of this as x to the 2/3 over 1. I multiply the top and bottom by 3x to the 1/3. So x to the 2/3 times 3x to the 1/3 over 3x to the 1/3. And that’s plus, remember over here I have 2x plus 8, over the same denominator; 3x to the 1/3.

Now when I multiply these here, I’m going to get x to the 2/3 times x to the 1/3. You add the exponents in a case like that. You get x to the 1, 3x to the 1. So this is 3x over 3x to the 1/3 that’s a 3, plus 2x minus 8 over the same denominator. And now I’m ready to combine these two. 3x plus 2x minus 8 is 5x minus 8. Over 3x to the 1/3.

Critical points. I have to look for two things; first where is the derivative equal to 0, and where is it undefined? And this time there is a place where the derivative is undefined. The derivative is undefined at x equals 0. I need to do a quick reality check.

Critical points have to be in the domain of a function. So x equals 0 will not count as a critical point, if it's not in the domain of the original function. It is in the domain of the original function. X equals 0 works fine here. So x equals 0 is in the domain and the derivative is undefined there.

So that counts as a critical point. So let me just make the note. So this is h(x), h'(0) is undefined. So x equals 0 is a critical point and I’ll abbreviate that c.p. And now looking back here again, I also need to find whether the derivative equals 0. The derivative equals 0, when the numerator equals 0.

So h'(x) equals o when 5x minus 8 equals 0, and that happens when x equals 8/5. And so this is another critical point. Let me just write that another critical point. And so my two critical points are x equals 0 when x equals 8/5. I was also asked to explain the geometric significance of this critical points. So I need to go back to the graph and show you that.

If you take a look at this graph, you can figure out what point we are talking about here. This is the point at x equals 0. Here the significance isn’t the same as before. Before we had a horizontal tangent at our critical point here. Here this would represent, because the derivative is undefined here, a vertical tangent. So imagine if you are trying to draw a tangent, a line that went the same direction as the curve, the best you could do is draw a vertical line here.

And that means that these slopes actually do go vertical when they meet at the point 0,0. So this blue line represents a vertical tangent. The other critical point is x equals 8/5, which I’m guessing is the x coordinate of this point 8/5. Because at that point, the derivative was 0. So at this point we do have a horizontal tangent.

So in some cases the critical point is going to represent a place where the graph has a horizontal tangent. But in some cases, it will represent a vertical tangent.