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Curve Sketching with Derivatives - Problem 1
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.
Let’s do a curve sketching problem. I’m asked to graph the function f(x) equals x³ minus 6x² minus 63x plus 42. Now it’s pretty easy to find the derivative. I did it before hand as a quick exercise here. I got first derivative is 3x² minus 12x minus 63. And I took the liberty of factoring that as well. It’s really important to do that because, we’re going to need to find to find the critical points.
Second derivative is 6x minus 12 and we’ll want to critical points for that derivative as well. Now the critical points are -3, and 7. The zeros of the first derivative. We’ll also want to examine what happens at x equals 2. That’s the zero of the second derivative, right in between. -3 and 7, zeros of the first derivative. I’m going to mark a zero here and here on the line of the first derivative. And then I want to test each of these 3 intervals to see if the derivative’s positive or negative.
Just really quickly looking at the factored form, if I try something to the left of -3 like -4, I’m going to get a negative number here and a negative number here and the product will be positive. So it will be positive here. If I try something between -3 and 7 like 2, I’ll get 5. -5, it will be negative. If I try something to the right of 7, like 8, that will be positive, that will be positive. Product of positive so it’s positive.
Now let’s go the second derivative. The second derivative is zero at 2, and if we look at this function, if I plug in a number to the left of 2 like 0, I’m going to get a negative value. It’s a linear function so it’s just going to go from negative to positive.
Let’s look at what we have here. Looks like our function is going to increase but be concave down in this interval. It’s going to decrease and be concave down over here. It will decrease and be concave up. And then it will increase and be concave up. So something like that.
So our key points are going to a local maximum at -3, a local minimum at 7 and at 2, looks like we have an inflection point. Goes from concave down to concave up. These will be some of our key points.
Let me just write the values here. We have f(-3), that turns out to be 100. I took the liberty of calculating that before hand. F(2) at the inflection point… I’m sorry this was 150. F(2) is -100 and f(7) is -350.
The only other point that I really need is, because it’s really easy, the y intercept is 42. So I could point out that f(0) is 42. So let me plot those points. -3 is about here, let’s plot 2 there and 7 about there. They’re actually equally spaced. So here’s -3, 2, 7. -3 I’ve got 150. I’m going to call that 150. And then 100 will be about 2/3 of that. So -100 is about here. And it’s a drop of 250 from 150 to -100, the same drop from here to here, so about down here.
Each of these points here at -3 and here at 7, we have a horizontal tangent. I’m going to draw a little horizontal line to remind myself to draw a horizontal tangent there. We have an inflection point at 2. To the left, remember it's increasing concave down, so it’s going to look something like this. To the right of -3 it’s