Calculus Help
- Thread starter pinky
- Start date
Pinky...are you sure you gave us the correct problem? The problem you gave us is (x+2x^(1/2))^(1/2) and I continue to get the same answer.
here is the link to the f'(x) and f''(x)
http://wims.unice.fr/wims/wims.cgi?session=MO4064D1BB.1&lang=en&cmd=reply&module=tool%2Fanalysis%2Ffunction.en&fn=%28x%2B2x%5E%281%2F2%29%29%5E%281%2F2%29&substitute=&show=deriv1&show=deriv2¢er=0&dev_order=5&ileft=&iright=&left=&right=&lower=&upper=&pleft=&pright=&num_precision=12&format=t
here is the link to the f'(x) and f''(x)
http://wims.unice.fr/wims/wims.cgi?session=MO4064D1BB.1&lang=en&cmd=reply&module=tool%2Fanalysis%2Ffunction.en&fn=%28x%2B2x%5E%281%2F2%29%29%5E%281%2F2%29&substitute=&show=deriv1&show=deriv2¢er=0&dev_order=5&ileft=&iright=&left=&right=&lower=&upper=&pleft=&pright=&num_precision=12&format=t
Yes, that is how the question is on the exam...I went to the site and got the same answer...the question wanted to know what the value of the second derivative was at X=1...which is negative when you put it in the equation...yet no possible answer to select is negative or close the the number that equation gives even if you ignored the negative sign
:shrug: :scared
:shrug: :scared
-0.2003877
-0.2003877
The first derivative at x = 1 is 0.49245589. The first derivative is value of the slope of the function at a given value of x.
The second derivative at x = 1 is -0.2003877. The value of the second derivative gives you the concavity of the function at the give value of x.
The original function is a square root function. The slope of the tangent line at x = 1 is 0.492, which means the function is increasing at that point. At that same x value, the second derivative is negative, which tells us that the function concaves down at that point.
What is the point of all this?
Calculus is not required to handicap sporting events but comes in handy in other professions.
Imagine this problem... A dairy farmer plans to fence in a rectangular pasture adjacent to a river. The pasture must contain 180,000 square meters in order to provide enough grass for the herd. What dimensions would require the least amount of fencing if no fencing is needed along the river?
This is an optimization problem that requires the first and second derivative of a function to solve and varify that the answer is minimized.
-0.2003877
The first derivative at x = 1 is 0.49245589. The first derivative is value of the slope of the function at a given value of x.
The second derivative at x = 1 is -0.2003877. The value of the second derivative gives you the concavity of the function at the give value of x.
The original function is a square root function. The slope of the tangent line at x = 1 is 0.492, which means the function is increasing at that point. At that same x value, the second derivative is negative, which tells us that the function concaves down at that point.
What is the point of all this?
Calculus is not required to handicap sporting events but comes in handy in other professions.
Imagine this problem... A dairy farmer plans to fence in a rectangular pasture adjacent to a river. The pasture must contain 180,000 square meters in order to provide enough grass for the herd. What dimensions would require the least amount of fencing if no fencing is needed along the river?
This is an optimization problem that requires the first and second derivative of a function to solve and varify that the answer is minimized.
I once had a pet turtle, I loved it very much. I had to keep it outside in a cooler and id go see it every day and night. One night I forgot to close the lid on the cooler and the turtle managed to escape and run away. I may have only spent little time with that turtle but I know memories will last a lifetime.
:mj07:
I had to throw that in there because thats what I wrote on my calculus test and took a 3%.
:mj07:
I had to throw that in there because thats what I wrote on my calculus test and took a 3%.
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On any calculus test...when in doubt, take the first derivative of the function.
1.5396 is the answer if you gave us the question properly and plug 1 in for x. i took calclulus a year ago and the shit isnt that hard. you just need to decrease the exponent by 1 and multiply the whole thing by the original exponent to get each derivative. the exponent of a sqrt is 1/2.
honestly i think you have may have written it wrong, because it isn't in good form. why would it have (Sqrt X + (SqrtX)) written like that. There woudl be no need for the extra parenthesis and it could be easily written as (2sqrt(x)). most time problems that the purpose isnt for you to simplify it are you usually given in their simplest form. are you sure you wrote it correctly?
There are the steps for what you gave
sqrt(x +(sqrt x + (sqrt X))....
I simplified it to
(x +2Sqrt(x))^1/2
1st deriv- since i have the whole equation raised to a power i can take a deriviative without messing with the inside. multiply whats in parenthesis by exponent 1/2 and decrease exponent by 1
1/2(x+2Sqrt(x))^-1/2
2nd Deriv -same process as first deriviative (multiply equation by -1/2 and decrease power by 1)
-1/4(x+2Sqrt(x))^-1 1/2
plug in 1 and i get that number above.
honestly i think you have may have written it wrong, because it isn't in good form. why would it have (Sqrt X + (SqrtX)) written like that. There woudl be no need for the extra parenthesis and it could be easily written as (2sqrt(x)). most time problems that the purpose isnt for you to simplify it are you usually given in their simplest form. are you sure you wrote it correctly?
There are the steps for what you gave
sqrt(x +(sqrt x + (sqrt X))....
I simplified it to
(x +2Sqrt(x))^1/2
1st deriv- since i have the whole equation raised to a power i can take a deriviative without messing with the inside. multiply whats in parenthesis by exponent 1/2 and decrease exponent by 1
1/2(x+2Sqrt(x))^-1/2
2nd Deriv -same process as first deriviative (multiply equation by -1/2 and decrease power by 1)
-1/4(x+2Sqrt(x))^-1 1/2
plug in 1 and i get that number above.
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Chain Rule
Chain Rule
PharoahUB, it's not as simple as making the exponent the coefficient, then decrease that exponent by 1 unit. You have to also take the derivative of the base, which in this case is using the chain rule twice.
The orginal function is a square root function. The parent graph of the square root function has positive slopes throughout, making the first derivative positive.
The second derivative is the slope of the graph of the first derivative. In the given function, the graph of the first derivative is a decreasing function. Therefore, the second derivative must be negative.
The first derivative of the given function can be found by pencil and paper, then with a TI-84 calculator, the second derivative can be found at x = 1.
Because of the chain rule, it would take at least an hour to take the second derivative of
the square root of (x + square root of (x + square root of (x)))
manually. I am sure Newton or Leibniz can do it, but for most of us, it could take a lifetime.
Chain Rule
PharoahUB, it's not as simple as making the exponent the coefficient, then decrease that exponent by 1 unit. You have to also take the derivative of the base, which in this case is using the chain rule twice.
The orginal function is a square root function. The parent graph of the square root function has positive slopes throughout, making the first derivative positive.
The second derivative is the slope of the graph of the first derivative. In the given function, the graph of the first derivative is a decreasing function. Therefore, the second derivative must be negative.
The first derivative of the given function can be found by pencil and paper, then with a TI-84 calculator, the second derivative can be found at x = 1.
Because of the chain rule, it would take at least an hour to take the second derivative of
the square root of (x + square root of (x + square root of (x)))
manually. I am sure Newton or Leibniz can do it, but for most of us, it could take a lifetime.
it is that easy. the base could easily be called X or Y or whatever the hell you want and manipulated the same way then plugged back in. lets make y=x+2sqrt(x)
now we have y^1/2
take that derivitive twice and its now -1/4y^-1.5
sub back what y equals too. plug in 1 for x.
now we have y^1/2
take that derivitive twice and its now -1/4y^-1.5
sub back what y equals too. plug in 1 for x.
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Ok I got a engineering major to solve this. Hope he got it right.
problem (x+2x^(1/2))^(1/2)
first derivative (1/2)(x+(2x^(1/2)))^(-1/2) * (1+x)
second derivative (-1/4)(x+(2x^(1/2)))^(-3/2) * (1+x)^2 + (1/2(x+(2x^(1/2)))^(-1/2)
sub 1 for x (-1)(1+2^(1/2))^(-3/2) + .5(1+2^(1/2))^(-1/2)
=.05521
I really want to know if this is the right answer or what the right answer is
problem (x+2x^(1/2))^(1/2)
first derivative (1/2)(x+(2x^(1/2)))^(-1/2) * (1+x)
second derivative (-1/4)(x+(2x^(1/2)))^(-3/2) * (1+x)^2 + (1/2(x+(2x^(1/2)))^(-1/2)
sub 1 for x (-1)(1+2^(1/2))^(-3/2) + .5(1+2^(1/2))^(-1/2)
=.05521
I really want to know if this is the right answer or what the right answer is
It all depends on the orginal problem. We are all interpreting the original function differently. If the original function is a square root function, then the first derivative at x = 1 is positive. The concavity, which is the second derivative, at x =1 must be negative.
Again, it's all how we view the original function. I see it as
the square root of (x + square root of (x + square root of (x)))
Again, it's all how we view the original function. I see it as
the square root of (x + square root of (x + square root of (x)))
I still don't think any of the answers are right...I scanned the question, but cannot put it on here...if anyone would like I could e-mail them the scanned question with all of the possible answers....or is there another way for me to share the question?
I give up...had another person try to help me do it and we got -.2511....
tell me when you get the answer
tell me when you get the answer
Get my email from Jack. Send me the orginal problem. We are all reading the original problem differently.I still don't think any of the answers are right...I scanned the question, but cannot put it on here...if anyone would like I could e-mail them the scanned question with all of the possible answers....or is there another way for me to share the question?
If the original function is a square root function, then the second derivative at x = 1 must be a negative value because the concavity at that point is opening down.
The website posted on page 1 of this thread will give you the first and second derivative if you enter the function. It doesn't, however, give you the value of the derivatives at x = 1. I use a TI-84 calculator to find those values.
Thanks for not giving up on me cryboy...I sent an e-mail to jack requesting your e-mail...I will send you the question with possible answers as soon as I get it from him.
