Download - Lesson 66 – Derivative Tests
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LESSON 66 – DERIVATIVE TESTSOctober 30, 2013Fernando Morales, Human Being
Calculus!
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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.
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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.
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Also find the local maxima and local minima points.
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Also find the local maxima and local minima points.
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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?
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Determine the intervals where the graph below concaves upward and the intervals where the graph concaves downward.
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Determine the points of inflection in the graph below.Concavity changes at points B, C, D, and P. Thus they are inflection points.
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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.
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Hence points of inflection at bothx = 0 and x = 2
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