Critical Points - Problem 2
Right now I’m solving problems that ask me to look for critical points of a function. And here is another example. Find the critical points of the function p(x) equals x² minus 3x at quantity times e to the minus 0.5x.
Now this is a product, so I’m going to have to use the product rule on the derivative. Remember, I need the derivative because critical points are points where the derivative equals 0 or is undefined. So I’m going to take the derivative and see where that happens.
And remember the product rule is the first times the derivative of the second. So I’m going to do x² times 3x times the derivative of e to the minus 0.5x. And that’s -5 e to the –0.5x, plus the second times the derivative of the first.
That’s e to the derivative of the 5x, times the derivative if the first that’s 2x minus 3. Now as always when I’m looking for critical points it's best to completely factor your answer. It's easier to find the 0’s of this derivative when you factor it. So let me observe that both of these terms have an e to the minus 0.5x.
In fact, both of them have a minus 0.5 e to the 0.5x, if I factor correctly over here. I’ll show you what I mean. Minus 0.5 e to the -0.5x. You can see that that factors out of these two term because it's already factored, x² minus 3x. But over here, I can turn this into a -0.5e to the -0.5 x. If I remember to multiply inside by the correct thing. For example to get 2 multiplying by -0.5 I would need -4. And so you see that -4 times –0.5 is 2. To get -3, I would have to have a plus 6. 6 times – 0.5 is -3.
So there I have managed to factor out a -0.5, and now I have exactly the same factor here and here. And that means I can combine –0.5e to the -0.5 x. And now I have x² plus 3x here, and minus 4x plus 6. So a total of x² minus 7x plus 6. Now hopefully this will factor as well because I want to find the critical points.
So –0.5e to the –0.5x. And I’m going to hope that it factors nicely with integer coefficients. And I think it does. If I use x minus 1, x minus 6, I will get this do multiply out of 6. And I get a minus x and a minus 6x minus 7x, that works perfectly.
So now my goal is, remember that it is the derivative. My goal is to figure out where this equals 0 and where it's undefined. So let me deal with the undefined part first. This is never undefined this part. A polynomial is never undefined. And this part an exponential function, exponential functions are defined for all x, so that’s fine.
Now let’s examine where it could possibly equal 0. So I set this equal to 0. Now again, an exponential function can never equal 0. E to any power always positive times the number now it's negative, but still not 0. This is never going to be 0.
Only these two factors can be 0. This 0 when x equals 1, I’ll write that one here. And this one is 0 when x equals 6, and these are my critical points.
So my critical points for this function turn out to be x equals 0 or x equals 6.