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Math Help - Decreasing Functions

  1. #1
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    Smile Decreasing Functions

    Hello!
    This is a question about the function f(x) = 1 + e2x.


    First I have to find f(x). I have found this to be: -2e^-2x


    Then I have to:
    Let P be the point on the graph of f where x = 1/2.
    (b) Find an expression in terms of e for

    (i) the y-coordinate of P;

    (ii) the gradient of the tangent to the curve at P.
    and finally I have to

    (c) Find the equation of the tangent to the curve at P, giving the answer in the form y = ax + b.

    I would appreciate your help.
    Thanks,
    Amine
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  2. #2
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    Quote Originally Posted by Aminekhadir View Post
    Hello!
    This is a question about the function f(x) = 1 + e–2x.


    First I have to find f(x). I have found this to be: -2e^-2x


    Then I have to:
    Let P be the point on the graph of f where x = –1/2.
    (b) Find an expression in terms of e for

    (i) the y-coordinate of P;

    (ii) the gradient of the tangent to the curve at P.
    and finally I have to

    (c) Find the equation of the tangent to the curve at P, giving the answer in the form y = ax + b.

    I would appreciate your help.
    Thanks,
    Amine
    For part b, just substitute x=-\frac{1}{2}, to obtain the function value at the given point. f(\frac{1}{2})=1+e^{-1}

    This is also the y-coordinate of P. So the point P is (\frac{1}{2},1+e^{-1}). This takes care of (i).

    The gradient of the curve at P is just the function value of the derivative at P. So f'(-\frac{1}{2}). I'll let you evaluate that to finish part (ii).

    For part (c) use the formula of a line:

    y-f(a) = f'(a)(x-a)

    Part (b) gave you all of the information you need to fill in this formula. Does this make sense?
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  3. #3
    MJ*
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    Quote Originally Posted by Aminekhadir View Post
    Hello!
    This is a question about the function f(x) = 1 + e2x.


    First I have to find f(x). I have found this to be: -2e^-2x


    Then I have to:
    Let P be the point on the graph of f where x = 1/2.
    (b) Find an expression in terms of e for

    (i) the y-coordinate of P;

    (ii) the gradient of the tangent to the curve at P.
    and finally I have to

    (c) Find the equation of the tangent to the curve at P, giving the answer in the form y = ax + b.




    Perhaps u have missed the power ^ in your question,

    For y cordinate of p,simply substitute x=-1/2 in your origional func;you will get corresponding value of y

    For gradient(slope) of tangent; you derivate the func like you have done and put x=-1/2;
    value of y=-2e will be your slope;

    For equation of tangent,just ut x1,y1 of point P(from part a) and m=slope of tangent (-2e)
    in the standard equation of line:

    (y-y1)=m(x-x1)

    and arrange your ans

    Hope that helps !!
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