So it just hit me that it's Monday and not Sunday...I hate that.
Anyway, I'm going to explain some AP questions that I got right! Yippee!!
Non-Calculator Portion:
1. If g(x) = 1/32x^4-5x^2, find g'(4).
First step: take the derivative
1/8x^3-10x
Second Step: Plug in 4
1/8(4)^3-10(4)
Third Step: Solve
Answer = -32
5. Evaluate lim h->0 [5(1/2+h)^4-5(1/2)^4]/h
First Step: This is a definition of a derivative, so you take the derivative of 5(1/2)^4
4(5)(1/2)^3
Second Step: Solve
Answer = 5/2
Calculator Portion:
34. The graph of y=x^3-2x^2-5x+2 has a local maximum at:
First Step: Plug into your calculator
Second step: 2nd->Calc
Third Step: go to maximum
Forth Step: enter your bounds and then a "guess"
Answer: (-.786,4.209)
39. Find two non-negative numberfs x and y whose sum is 100 and for which (x^2)(y) is a maximum.
First step: Add up answer choices to make sure all equal 100
Second Step: Plug the x and y values into the given formula and see which is the highest number
Answer: x=66.667 and y=33.333
QUESTIONS:
How do you find the domain of a function without a calculator? I've always had trouble with domain and range even in Advanced Math.
How do you work piecewises? Number 7 is an example on the Non-calculator portion: Find k so that f(x) {(x^2-16)/x-4; x *does not equal* 4 and k; x=4 is continuous for all x. a) all real values, b)0, c)16, d)8, e) no real values
How do you take the derivative of a trig function squared? (sin^2(x) or cos^2(x))
Can someone go over the Mean Value Theorem for Derivatives? Number 12: Find a positive value c, for x, that satisfies the conclusion of the MVTD for f(x)=3x^2-5x+1 on [2,5] a)1, b)13/6, c)11/6, d)23/6, e)7/2
Hope everyone enjoys their day off!
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To take the domain like the one we had on the test, you set the inside of the square root equal to zero and solve for x. Then you set up intervals between the x values and plug them into the original function to see which one gives you a positive value. For the Mean Value Theorem ones, you use the formula f(b)-f(a)/b-a to find the slope. Then you take the derivative of the function and set it equal to the slope to find the answer. Hope this helps girl!
ReplyDeletemean value theorem is f(b)-f(a) divided by b-a
ReplyDeleteif you have [2,5] then b=5 and a=2
plug in 2 and 5 to the equation given:
F(b)=51 & F(a)=3
so it's 51-3 divided by 5-2
which is 48/3 which is 16
then you take that answer and set it equal to the derivative of the equation given to you which is 6x-5
so that gives you 16=6x-5 so solve for x
x=3.5 which is 7/2 so the answer is E
To take the derivative of a trig function squared, like sin^2(x) for example...it's really easy.
ReplyDeleteWhat I like to do is visualize it like this
(sin(x))^2
Now it's just like if you had anything else in parenthesis. First you bring the 2 to the front, so you have
2 (sin(x))^1
Now, you times by the derivative of the inside of the parenthesis (chain rule).
So we get
2sin(x)cos(x) as our final answer.
to take the derivative of a trig function squared is quite easy. work from the outside in..
ReplyDeleteso start with the exponent.
cos^2(x) = 2cos(x)
then go to the trig function itself
2cos(x) = 2-sin(x)
therefore answer is -2sin(x)
work from the outside to the inside. you will end up w/ -2sin(x)
ReplyDeleteRicky and Sarah see John's comment. He is right your answers are not correct. The trig function doesn't disappear. We have also worked this together on several of the APs that we did solutions together in class to.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteyou can't just work on part of the derivative, when there's a trig function squared always remeber that you're going to be doing two thingse, just like john said, you're using the chain rule. after moving thw two to the front, you have to recognize that you still haven't done anything with the trig function.. derivative of sin is cos.. just take a little more time to think about what the problem's asking you to do and it'll all make sense.
ReplyDeleteWhat year is this test?
ReplyDeleteWhat year is this test?
ReplyDelete