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Intervals of Increase and Decrease - Problem 2
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Remember that a function is decreasing when its derivative if negative. So, to find where a function decreases, first calculate its derivative. In order to find where the derivative is negative, first set it equal to 0 and solve for x.
Pick a number between each pair of values, plug it into the derivative, and check if it is positive or negative. Check this with values on each interval. For the smallest solution of x, check this with any number that is less than that solution, and for the largest solution, check this any number greater than that it. For example, if the derivative of a function is 0 when x=-3, x=1, and x=17, evaluate the derivative in four places: when x<-3, when -3
If the value of the derivative at a point on an interval is negative, then your function is decreasing.
Let’s do a problem that involves finding where a function increases or decreases. So I have a problem that says find the intervals on which f(x) equals 4x cubed minus 45x², minus 48x plus 60 decreases.
By the increase in decreasing test, I need to look at where the derivative is negative. Because that will tell us where the function decreases. So where is f'(x) negative less than 0? So the first thing I want to do is find the derivative of this function. So let’s work up here.
f'(x) and we’ve got 4x cubed, the derivative of that is 12x². And we’ve got -45x², the derivative of that is -90x. The next term, the derivative of -48x is going to be -48 and the derivative of 60 is 0. So I get -48. And now I have to determine whether this quadratic is positive or negative. Actually I need to know to know where it's negative specifically to find out where f(x) decreases.
So I’m solving this inequality. Now when you solve a polynomial inequality you want to completely factor the polynomial. So let me factor this completely and note that each of these constants has a common factor of 6.
So I’ll pull this 6 out and I’m left with 2x² minus 15x minus 8. And then I can factor this quadratic. It will be 2x plus or minus something and x plus or minus something. And the last two numbers have to be factors of 8. So I could pick 2 and 4 or 1 and 8. Turns out 1 and 8 will work nicely because 8 times 2 is 16. 1 times x is going to be 1x. So 16x minus x is 15x. So if I put my -8 here, and my plus 1 here, that will work perfectly.
-16x plus x is -15x. The trick to figuring out exactly where this quantity is going to be negative, is to draw a sign chart. And a sign chart is just a number line and on the number line you want to label points where this thing equals 0.
You could see that it's 0 at -1/2 and at 8. And this does not need to be to any kind of scale. You just need to put the numbers in their correct relative positions. So the derivative we’ll put over here is 0 and to these points. We want to find whether it's positive or negative here, here and here. And all I need to do is test one point in each of those 3 intervals.
So let’s start with this left hand interval. A point to the left of -1/2 would be -1. If I plug -1 in here, I get -2 plus 1 I get a positive number. I don’t need to know what it is, I just need to know that it's positive. -2 plus1 is -1 so that’s negative. And then -1 minus 8 is -9 also negative. So I have a product of choose two negatives which will be positive. And then I’ll try a number between these two.
So 0 would be good. Pick easy numbers if you can. 0 is an easy number so 2 times 0 plus 1 gives me 1, which is positive. 0 minus 8 is -8 negative. So I have positive times negative I’m going to get a negative.
And finally over here, let’s try 9. 2 times 9 plus 1 is 19, that’s positive and 9 minus 8 is 1 also positive. So this is a product of positive numbers which will be positive. Now remember I want to find out where this quantity is negative. So here’s is where I focus; between -1/2 and 8. So on this interval f'(x) is less than 0 and that means that on this interval f(x) decreases.
And that’s the answer to our problem. Where does f(x) decrease? On the interval between -1/2 and 8.
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