 ###### 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 - Concept

# Intervals of Increase and Decrease - 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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Remember that a function is increasing when its derivative if positive. So, to find where a function increases, first calculate its derivative. In order to find where the derivative is positive, 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 -317. You can pick any number between the points you are looking at.

If the value of the derivative at a point on an interval is positive, then your function is increasing.

Let’s do another problem that involves looking for where a function increases or decreases. So here my function is f(x) equals the quantity x² minus 3 times e to the x. I want to find the intervals on which this function increases.

So by the increasing decreasing test, I would look for where f'(x) is positive. So this is my question where is f'(x) greater than 0 or positive? So I’m going to need to take the derivative of this function.

f(x) is x² minus 3 times e to the x. Now this is a product of two functions, so my derivatives is going to need the product rule. So it’s going to be first times the derivative of the second. Now the first is x² minus 3 the derivative of the second function e to the x is just e to the x. Plus the second e to the x, times the derivative of the first, which is 2x. And I can combine these two into a single term if I factor out the e to the x. So I get x² plus 2x minus 3 from here times e to the x.

And finally let me factor this one little bit more. This could be factored intro x plus or something x or minus something e to the x. I’m going to want these last numbers to be factors of 3. So 1 and 3 would work.

And I noticed that I needed 2x here so I’m going to be plus 3 and minus 1. That will give me my -3 here. It will give me 3x minus x 2x here. Perfect now I want to find out where this is positive. And so once again I’m going to use a sign chart. On the sign chart, this is the sign chart for f'(x).

On the sign chart I want to label points where this derivative is 0. Now it will be 0 when x is -3, it will be 0 when x is 1. But this factor does not contribute any 0’s, in fact e to the x is always positive. So we don’t really have to worry about this factor contributing to the solution set. This is just going to be positive all the time.

So we put 0 here and here. And then let me test points in each of these three intervals to determine whether the derivative is positive or negative. So let me try -4. -4 will give me a -1 here, a -5 here two negatives, and this is positive. So I get a positive. And then let me try something between -3 and 1. I would like to pick 0 if I can. So 0 plus 3 is positive, 0 minus 1 negative, positive. 1 negative, so this is going to be negative.

And finally I’ll try 2 in this interval 2 plus 3 is 5 positive 2 minus 1 is 1 positive. And this is always positive so positive. And so we see that f' is positive here for x less than -3, and here for x greater than 1.

That’s where f' is positive. So on these two intervals f' is positive. And so that means on these two intervals, f(x) increases. So the answer to our question where does f(x) increase? On the interval of x less than -3, and on the interval x greater than 1.