# Intervals of Increase and Decrease - Problem 1

A function is increasing on an interval if its slope is positive on that interval, and decreasing if its slope is negative on that interval. When given an interval, simply plug in a value of x from somewhere in that interval to determine if the slope is positive or negative. For example, if you are looking at a function for x between -3 and 7, pick a number between -3 and 7, and plug it into the derivative. If the answer is positive, the function is increasing between -3 and 7, and if the answer is negative, the function is decreasing between -3 and 7.

When f'(x)=0, the function is neither increasing nor decreasing.

So we just introduced the increase and deceasing test. This is a test that says if f' for a function is positive on the interval, then f increases on that interval. If f' is negative on an interval then f decreases on the interval.

We can use that to sketch the graph of a function if we have some information about where f is positive and where it's negative. So let’s take a look at this example. It says draw a possible graph of y equals f of x, if f'(x) is positive for x between -1 and 5

f'(x) is negative for x less than negative 1, and x greater than 5. And if f'(x) is 0 at x equals -1 and x equals 5. So it looks like the important numbers here are -1 and 5. So I’m going to mark those on a graph really quickly.

I’ll make this -1 and this 5. Exact spacing doesn’t matter here, I just want to get the qualitative details of where the function is increasing and where it's decreasing. Now I’m told that the derivative is 0 at -1 and 5 to the left of -1. So for x less than -1, the derivative is negative and that means that the function has to decrease.

So it’s got to decrease to the left. So it's going to come in from above. And then it’s going to have a 0 derivative at -1. And that means the tangent there has to be horizontal. And then between -1 and 5, we are told f'(x) is positive. So that means that the function increases.

So from -1 to 5 it increases and it finishes off at x equals 5 by having f'(x) be 0. So we have to have a horizontal tangent here as well. And finally for x greater than 5, the function’s got a negative

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