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Math Help - Derivatives, Abs. Extrema, Implicit Diff.

  1. #1
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    Derivatives, Abs. Extrema, Implicit Diff.

    Hello guys, I have multiple questions today regarding several problems I have had problems working out. I will post the problem following how I have attempted to work it out. I have an exam on the topics of derivatives, absolute/relative extremas, implicit diff, relative rates later this week, so if you have any tips regarding these topics please share them.

    1.) Find the derivative.
    y = 23 ^ -x, I thought the answer to this would be ln(23)(23^-x), however in the answer key it states it as -23^-x, how is this so?

    2.) 2nd Derivative of function f(x) = x/x+1, is the answer the same as the function itself? F''(x) = x/x+1?

    3.) Abs. Extrema
    f(x) = 1/x+2 [-4,1]
    f'(x) = -x-2/(x+2)^2 ? x=0 x=-2, (0 , 1/2) ??? Is this how this is suppose to be worked out?

    4.) Abs Extrema
    f(x) = x^(4/3) - x ^(2/3)
    f'(x) = 4/3x^1/3 - 2/3x^1/3, where do I go from here to solve this? I am confused as to how I would make some of these equations =0, any tips would be nice.

    5--The last problem I have is a word problem related to business. I am not sure what it is exactly asking for in the problem however it states that there is a off-load oil docking facility 5 miles off shore. the nearest refinery is 8 miles easy of the facility. a pipe would cost 200k on land and 300k in water. Locate point B to minimize the cost. ------- So basically, there is a triangle drawn to show where the facility is and where the refinery is(on land), point B is suppose to be on land halfway from point A and the refinery. point A is the point on land that is 5 MILES from the facility off shore. Here I will just type out the problem to make more sense, I need more help with the problems above, this is just an extra however it will not be heavily tested.

    Supertankers off-load oil at a docking facility shore point 5 miles offshore. The nearest refinery is 8 miles east of the docking facility a pipeline must be constructed connecting the docking facility with the refinery. the pipeline costs 300,000 per mile if constructed underwater and 200000 per mile if over land. Locate point B to minimize the cost of construction.
    A triangle is drawn to show the facility point A(5 mile from docking) point B(from docking to a horizontal distance towards the refinery however not quite there.)


    Thanks for all the help, I hope this post was not quite long however I waited until I was done with all of the problems before pointing out the ones I needed help with. Once again, thank you.
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  2. #2
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    Any help works, on certain problems, still having problems figuring out how they work.
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  3. #3
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    Anyone?
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  4. #4
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    Quote Originally Posted by cokeclassic View Post
    Hello guys, I have multiple questions today regarding several problems I have had problems working out. I will post the problem following how I have attempted to work it out. I have an exam on the topics of derivatives, absolute/relative extremas, implicit diff, relative rates later this week, so if you have any tips regarding these topics please share them.

    1.) Find the derivative.
    y = 23 ^ -x, I thought the answer to this would be ln(23)(23^-x), however in the answer key it states it as -23^-x, how is this so?
    I presume you mean the answer was given as -ln(23) 23^{-x}, since just " -23^{-x}" is not correct. The "-" comes from using the chain rule. The derivative of 23^x is ln(23) 23^x but since the exponent is "-x" you have to multiply by the derivative of -x which is -1.

    2.) 2nd Derivative of function f(x) = x/x+1, is the answer the same as the function itself? F''(x) = x/x+1?
    No, why would you think so? By the quotient rule, the derivative of f is f'(x)= \frac{1(x+1)- x(1)}{(x+1)^2}= \frac{1}{(x+1)^2} and then, using the quotient rule again, f"(x)= \frac{0(x+1)- 1(1)}{(x+1)^4}= \frac{x}{(x+1)^4}. Did you forget to square in the denominator?

    3.) Abs. Extrema
    f(x) = 1/x+2 [-4,1]
    f'(x) = -x-2/(x+2)^2 ? x=0 x=-2, (0 , 1/2) ??? Is this how this is suppose to be worked out?
    First, your derivative is wrong. (I am assuming that you mean f(x)= 1/(x+2), not (1/x)+ 2 which, strictly speaking is what you wrote.) f'(x)= \frac{0(x+2)- 1(1)}{(x+2)^2}= \frac{-1}{x+2}.
    An extermum occurs at one of three kinds of places.
    (1) Where the derivative does not exist, which does not happen here.
    This derivative does not exist at x=-2 but neither does the function value so we can ignore that.

    (2) Where the derivative is 0. Since a fraction is 0 only where the numerator is 0, and this numerator, -1, is never 0, that alos does not apply.

    (3) At the endpoints. f(-4)= 1/(-4+2)= -1/2 and f(1)= 1/3. The larger of those is 1/3 so the absolute maximum is 1/3 which happens at x= 1.

    4.) Abs Extrema
    f(x) = x^(4/3) - x ^(2/3)
    f'(x) = 4/3x^1/3 - 2/3x^1/3, where do I go from here to solve this? I am confused as to how I would make some of these equations =0, any tips would be nice.
    Once again, you have the derivative wrong. The derivative of x^n is nx^{n-1} and 2/3- 1= -1/3, not 1/3.

    f'(x)= (4/3)x^{1/3}- (2/3)x^{-1/3}= 0 is the same as (4/3)x^{1/3}= (2/3)x^{-1/3}. Now multiply both sides of the equation by 3x^{1/3} and divide by 2: 2x^{1/3+ 1/3}= 1 so x^{2/3}= 1/3. You solve that by taking the 3/2 power of both sides or, same thing, by taking the square root and then cube of both sides: x= \pm\sqrt{1/27}.

    5--The last problem I have is a word problem related to business. I am not sure what it is exactly asking for in the problem however it states that there is a off-load oil docking facility 5 miles off shore. the nearest refinery is 8 miles easy of the facility. a pipe would cost 200k on land and 300k in water. Locate point B to minimize the cost. ------- So basically, there is a triangle drawn to show where the facility is and where the refinery is(on land), point B is suppose to be on land halfway from point A and the refinery. point A is the point on land that is 5 MILES from the facility off shore. Here I will just type out the problem to make more sense, I need more help with the problems above, this is just an extra however it will not be heavily tested.

    A triangle is drawn to show the facility point A(5 mile from docking) point B(from docking to a horizontal distance towards the refinery however not quite there.)
    First I am going to assume that you mean that the pipe would cost 200k per mile on land and 300k per mile off shore. Also B is NOT 'halfway' between A and the refinary- it is between them and the problem is to determin how far from A to B.

    Let "x" be the distance from A to B- that will give you a right triangle with one leg "x" and the other 5. By the Pythagorean theorem, the straight line distance to the point where the pipe hits the coast is \sqrt{x^2+ 25} miles. Since building the pipe off shore costs 300k per mile the cost of building that portion is 300000\sqrt{x^2+ 25}. Since B is already distance x from A to the refinery, the distance remaining on land from B to the refinery is 8- x and the cost of building that will be 200000(8- x).

    The total cost of the pipe line would be C(x)=  300000\sqrt{x^2+ 25}+ 200000(8- x). That is what you want to minimize. You will want to differentiate it. You will probably find it simplest to write it as C(x)= 300(x^2+ 25)^{1/2} + 200000(8- x).




    Thanks for all the help, I hope this post was not quite long however I waited until I was done with all of the problems before pointing out the ones I needed help with. Once again, thank you.
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  5. #5
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    <br /> <br />
f"(x)= \frac{0(x+1)- 1(1)}{(x+1)^4}= \frac{x}{(x+1)^4}<br />
    for this part, you took the derivative of 1/(x+1)^2, however I dont understand, what exactly do you do with the (x+1)^2? would it not be 2(x+1)(1)?
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