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Critical Points - Problem 1
Critical points of a function are where the derivative is 0 or undefined. To find critical points of a function, first calculate the derivative.
The next step is to find where the derivative is 0 or undefined. Recall that a rational function is 0 when its numerator is 0, and is undefined when its denominator is 0. So, when looking at the derivative of the function, find the zeros of its numerator and denominator to find the values of x where the derivative is 0 or undefined. These values of x are the critical points.
Let’s do a problem. Find the critical points of the function r of x equals x² minus 5x plus 4 over x² plus 4. This is a rational function, so to take its derivative, I’m going to want to use the quotient rule.
So I’m looking for the derivative because, remember, the critical points are points where the derivative equals 0 or is undefined. And I can find those points by examining the derivative. And remember the quotient rule; it's low d high. So x² plus 4x, it's the low part. D high is the derivative of the upper part. That will be 2x minus 5 minus high d low x² minus 5x plus 4. Times the derivative of the low part 2x over the square of what's below.
And that means square the denominator. Now let me simplify this, I want to simplify and factor as much as possible that will make it easier to find critical points. Now looking at the numerator, it's not as bad as it looks. We have a cubic here and a cubic here.
Let me multiply this through. I want to get 2x times x² that’s 2x cubed, and then what I like to do is I like to look for decreasing powers of x. The x² term will come from minus 5 times x² minus 5x².
Then we’ll have plus 8x and then I’ll have minus 20. That’s this term. Now this term is going to give me 2x times x² minus 2x cubed. Then I’m going to have minus minus, so plus, 10x², and then I’m going to have minus 4 times 2x minus 8x. All that over x² plus 4 the quantity squared.
So I have some cancellation I can do. The 2x cubes cancel. Which is nice and then the 8x’s also cancel and I’m left with. Minus 5x² plus 10² that’s plus 5² minus 20. Let me rewrite that over here.
So this is where I have my R' so far. 5x² on top minus 20, all over x² plus 4 the quantity squared. This is my simplified derivative. Now remember my goal here is to find critical points. So I want to find where this derivative equals 0 or is undefined.
Let me tackle the undefined part first. Because I have a rational derivative, it's possible to be undefined. But it's only going to be undefined if the denominator can never equals to 0, which it can’t. X² plus 4 can’t equal 0. The smallest it's ever going to be is 4 and when you square that, 16. So it's not going ever to be 0. But the numerator can equal 0 and that means that our derivative could equal 0.
R'(x) equals 0 when the numerator equals 0. 5x² minus 20 equals 0. And so I just have to solve this equation for x. I get 5x² equals 20, x² equals 4, and then x equals plus or minus 2. These are my critical points.
So my critical points turn out to be x equals plus or minus 2.