I'm always confused with first and second derivative test. So I decided to relearn and hopefully explaining it will help me too.
First Derivative Test:
Example:
Find the critical pts of f(x)=(x^2-4)^(2/3)
(2/3)(x^2-4)^(-1/3)
4x/3(x^2-4)^(1/3) =0
2x=0
4x=0
x= 0, +2, -2 <---critical points
Now set up intervals and plug in numbers in between them. You are plugging into the FIRST DERIVATIVE. You don't really have to know the number, but you need to know if it will be positive (+ve) or negative (-ve) in order to know if it is decreasing or increasing.
(-infinity,-2) (-2,0) (0,2) (2,infinity)
f^1(-3)=4(-3)/3(9-4)^1/3 = -ve/+ve = DECREASING
f^1(-1)= -ve/-ve = INCREASING
f^1(1)= +ve/-ve = DECREASING
f^1(3)= +ve/+ve = INCREASING
Mins and Maxs?
x= -2, 2 <--mins
x= 0 <--max
Second Derivative Test:
You take the derivative and the second derivative. The values you get are NOT critical values; they are more like points of interest. Now when you plug in this time you must plug into the SECOND DERIVATIVE. This helps tell about concave up (+ve) and concave down (-ve).
6/x^2+3
1st deriv= -12x/(x^2+3)^2
2nd deriv= (36(x+1)(x-1))/(x^2+3)^3
x= +1, -1 <--not critical values
(-infinity,-1) (-1,1) (1,infinity)
f^(double prime)(-2)= +ve = concave up
f^(double prime)(0)= -ve = concave down
f^(double prime)(2)= +ve = concave up
pts of inflection x= -1, 1
My question is about the shortcut..I still don't understand how to do it and why we do that.
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The shortcut is very easy. It's main use is for multiple-choice only because in short answer you can't show the shortcut method for your work.
ReplyDelete1. You find the first derivative and set it equal to zero.
2. Solve for x.
3. Find second derivative and plug the x values into it. The biggest number is your absolute max. and the smallest number is your min.
You find the first derivative and set it equal to zero then Solve for x. and finally Find second derivative and plug the x values into it. The biggest number is your absolute max. and the smallest number is your min.
ReplyDeleteThe shortcut is pretty easy but remember you can't always use it because you need work for free response questions. First find the first derivative. Set it equal to zero and solve for x. Then find the second derivative. Plug the x values you found into it. The largest number is the absolute max while the smallest number is the absolute min.
ReplyDelete