This week, since we did not go over anything new, I am going to show you how to use the Second Derivative Test to find all possible points of inflection and intervals of concavity. Remember, points of inflection only happen where there is a change of concavity.
Example: f'(x)= 6/(x^(2)+3)
First, you have to take the derivative of that, and you have to use the quotient rule, so the beginning of the problem will look like, [(x^(2)+3)(0)-6(2x)]/(x^(2)+3)^(2), which simplifies to -12x/(x^(2)+3)^2 remember, that was just the first derivative.
Second step is to take the derivative of the first derivative, that would make this step called taking the second derivative.
Once again u need to use the quotient rule, so f''(x)={(x2+3)^(2) -(12)-[(-12x)2(x^(2)+3)2x} all that over (x^(2)+3)^4 then you get a bunch of stuff, then you simplify, then you cancel, and if I typed all that up it would be ridiculous, so I'm just going to type the end answer of the second derivative. Which is, (3)(6)(x+1)(x-1) all over (x^(2)+3)^3
The possible points of inflection are found in the numerator of the finished second derivative, in this case, if you look, it would be x=1, and x=-1
so then you set up your points, (-infinity, -1) u (-1, 1) u (1, infinity)
then you plug in. f''(-2)= positive value f''(0)=negative value f''(2)=positive value
then you know that your intervals concave up @ (-infinity, -1) u (1,infinity) or x<1,>1
and it is concave down @ (-1,1) or -1
and you're points of inflection are x=-1, and x=1
okay, my question comes from a quiz that we had, it was the last problem and I didn't know how to do it, the problem goes like this: The sum of the perimeters of two squares is 24. Find the dimensions of the squares that produce the minimum total area.
I know it's a optimization problem, but I just don't know how to solve it. Help would be appreciated.
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ok for that optimization question you have...the number four on the test we just got back.....
ReplyDeleteso we know that we have a perimeter of 24 right...well to make your secondary and primary equation you make two squares...we know that the area of a square is lenth times width! and that there is two lengths and two widths for each square so your seconday equation will be 2(2L+2w)...that's your perimater because you're adding them...and since we know that the sum of the two perimeters is 24 you set 2(2L+2w) equal to 0!
so now...dealing with optimization:
take your secondary equation and solve for...lets go with length
2[2(L)+2(w)] = 0
24=4(L)+4(w)
24-4(w)=4(L)
(L)=6-w
so now plug in L to the primary equation which is your area...so that's
A=(L)(w)
A=(6-w)(w)
A=6w-(w)^2
TAKE THE DERIVATIVE OF THAT:
A'=6-2w
-2w=-6
w=3
SO NOW TAKE WHAT YOU KNOW....W=3....and PLUG IT INTO THE SECONDARY FORMULA:
24=2(2(L)+2(w)
24=2(2(L)+2(3)
24=4L+12
12=4L
L=3
so knowing your W is 3 and your L is 3
[it's a square so they had to be the same]
you know that the width is three and the length is three!!
make sure you know which equations you need to use. like your primary and secondary
ReplyDeleteOptimization
ReplyDelete1 Identify primary and secondary equations your primary is the one your or maximizing or minimizing and your secondary is the other equation
2. Solve for your secondary variable and plug into your primary equation if your primary only has on variable this isn’t necessary
3. Plug into secondary equation to find the other value check your end points if necessary