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Math Help - advanced functions intro calc questions

  1. #1
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    advanced functions intro calc questions

    ok so im very new to calculus and just started learning limits...were exiting logs and entering limits (im a grade 12 student) my teacher has given me 2 question that have got me stumped.. here they are if anyone can help me out pllleaase help thanks

    solve for x

    2^x - 2^-x =4

    2)determine the equation of both lines that are tangent to the graph of
    f( x)=x^2
    and pass through point
    (1 , -3)
    Must use
    lim F( x ) - F ( a ) / x - a
    x->a
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  2. #2
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    Quote Originally Posted by jamman790 View Post
    ok so im very new to calculus and just started learning limits...were exiting logs and entering limits (im a grade 12 student) my teacher has given me 2 question that have got me stumped.. here they are if anyone can help me out pllleaase help thanks

    1) solve for x

    2^x - 2^-x =4

    2)determine the equation of both lines that are tangent to the graph of
    f( x)=x^2
    and pass through point
    (1 , -3)
    Must use
    lim F( x ) - F ( a ) / x - a
    x->a
    to #1):

    2^x-2^{-x} = 4~\implies~(2^x)^2-1=4\cdot 2^x~\implies~2^{2x}-4 \cdot 2^x-1=0

    This is a quadratic in 2^x . Use the substitution y = 2^x
    Solve for y
    Re-substitute to calculate the value(s) of x

    to #2)

    Let T(t, t^2) denote the tangent point on the graph of f.

    The slope of the tangent is m_t=f'(t) = 2t

    The equation of the tangent in T is:

    y-t^2=2t(x-t)~\implies~\boxed{y=2tx-t^2}

    This tangent has to pass through P(1, -3) that means the coordinates of P have to satisfy the equation of the tangent:

    -3=2t\cdot 1-t^2~\implies~t^2-2t-3=0~\implies~t=3~\vee~t=-1

    You get two tangent points and two tangents:

    T_1(-1,1) with tangent y = -2x-1

    T_2(3, 9) with tangent y = 6x-9
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  3. #3
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    can u please explain for the 1st question you awnsered (x^2 - X^-2 = 4) how you get to the second step or more specifically how u ended up with x^2x - 1? thank you
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  4. #4
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    Quote Originally Posted by jamman790 View Post
    can u please explain for the 1st question you awnsered (x^2 - X^-2 = 4) how you get to the second step or more specifically how u ended up with x^2x - 1? thank you
    He doesn't. I think you are misreading " (2^x)^2- 1= 4(2^x)"
    From 2^x- 2^{-x}= 4, multiply both sides of the equation by 2^x: (2^x)(2^x)- (2^x)(2^{-x})= (2^x)(4) which is the same as (2^x)^2- 1= 4(2^x). Now let y= 2^x and that equation becomes y^2- 1= 4y or y^2- 4y- 1= 0.

    Myself, I would have made that substitution to begin with: Let y= 2^x in the equation 2^x- 2^{-x}= 4 so that you have y- \frac{1}{y}= 4 and multiply on both sides by y: y^2- 1= 4y, which is, of course, exactly the same thing.
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  5. #5
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    OooooH! Thank you so much thanks guys, life saver
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