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1. Sketch the curves y=e^x and y=e^-2x, using the same axes. The line y=4 intersects the first curve at A and the second curve at B. Calculate the length AB to two decimal places.

2. Find the coordinates of the turning point on the curve y=2e^3x+8e^-3x and determine the nature of this turning point.

For 1, I have sketched the curves, but i'm not sure how to find the length. I thought I could do y=y to find the x co-ordinates (and use those as my limits) for each but I don't know how to find them out with e^x.

For 2, I differentiated it to get y=6e^3x-24e^-3x and don't know what to do after that.

Please help. Thanks in advance for any help!

  • Maths -

    2. find each x by

    ln 4= x so the point is ln4,4
    and the other point is
    x=-ln4/2 and the boint B is -1/2 ln4,4

    Use the distance formula to find the length.

  • MATH -

    1. Since the y coordinates of A and B are the same, the distance between the points will be the difference in the X values of those two points.

    2. The derivative is
    y' = 6e^3x -24e^-3x
    When that equals zero,
    e^3x = 4e^-3x
    e^6x = 4
    6x = ln 4 = 1.3863
    x = 0.23105 is an extreme point of the function. The value of y there is
    2 e^0.693 + 8 e^-0.693 = 2*2 + 8/(1/2) = 20
    Take the second derivative of y(x) to see if (0.23105,20) is a minumum or maximum. It is going to be positive. You also see that, since y becomes infinite as x gets much larger, it must be a minimum.

    y''(x) =

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