LESSON 66 – DERIVATIVE TESTSOctober 30, 2013Fernando Morales, Human Being

Calculus!

Between A and B the tangent lines have positive slope so f ’(x) > 0.Between C and D, the tangent lines have positive slope and so f ’(x) > 0.Between B and C, the tangent lines have negative slope and so f ’(x) < 0.

What type of slope is there between each of the intervals?

What does the sign of the derivative tell us about the function?It appears that f increases when f ’(x) is positive anddecreases when f ’(x) is negative.

Using the First Derivative Test:Step 1: Take the first derivative of the function.

Step 2: Find the critical points, that is find the values of x where f ’(x) = 0 and where f ’(x) does not exist.

Step 3: Determine the behaviour of f ’(x), in other words the sign of f ’(x) whether positive or negative at intervals whose endpoints are the critical points.

Step 4:If there is a sign change of f ’(x) between intervals at the critical point,then the critical point is either a local maxima or local minima.

Also find the local maxima and local minima points.

Also find the local maxima and local minima points.

Also find the local maxima and local minima points.

The two graphs above are both increasing function on the interval (a, b).Both graphs join point A to point B but they look different.How can we distinguish between these two types of behaviour?

Determine the intervals where the graph below concaves upward and the intervals where the graph concaves downward.

Determine the points of inflection in the graph below.Concavity changes at points B, C, D, and P. Thus they are inflection points.

Using the Second Derivative Test:Step 1: Take the first and second derivative of the function

Step 2: Find the critical points, that is find the values of x where f ’(x) = 0 and where f ’(x) does not exist. And evaluate f ’’(x) at the critical points.

Step 3: If f ’’(x) > 0 then there is a local maximum.If f ’’(x) < 0 then there is a local minimum.If f ’’(x) = 0, then you need further investigation. (Use first derivative Test)

Step 4:Apply the concavity test, by determining the sign of f ’’(x) between intervalswhere f ’’(x) = 0 and where f ’’(x) does not exist.If f ’’(x) > 0, then concave upwards.If f ’’(x) < 0, then concave downwards.

Step 5:Determine whether points of inflection exist, that is, if there is a sign change of f ’’(x) between critical points.

Hence points of inflection at bothx = 0 and x = 2