So this week in calculus it was a mixture between power point presentations and working on our study guides. One thing I understand in our study guide is tangent lines. All you do is take the derivative of the equation. Then plug in the x value from the point given giving you the slope. Next plug the slope and point in to slope intercept form which is y-y1=slope(x-x1). Also I understand some limit stuff. To find the vertical asymtote just set the bottom equal to 0. Also I understand how to take derivatives. Whether it be the easy ones where its just an equation, to product rule, or quotient rule. All the work and practice we did on it helped me out a lot and now I completely understand how to take the derivative of many equations.
But I still do not fully understand optimization. I just can not seem to get it fully at all. Also I do not really remember the Intermediate Value Theorem. I probably know how to do it I just need a refreshement. So if anyone can help me with either of these it would be great.
ok for the optimization questions you have...lets look at number four on the test we just got back.....[3.7 Homework Quiz..it was out of 20 points]
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!!
A problem should be something like this:
ReplyDeleteFind the value of c guarenteed by the intermediate value theorem f(x)=x^2-2x-3,[4,8], f(c)=12.
And the choices are A) 3 B) 2 C) 5 D) 7 E) 6
f(x)=x^2-2x-3
Plug in each answer choice to original equation and solve.
=5^2-2(5)-3 =12
The one that gives you f(c)=12 is the answer, C) 5.