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Thread: Intersection of tangent lines

  1. #1
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    Intersection of tangent lines

    1. f(x)= 2^x^2
    Find the point of the intersection of the tangent lines to f(x) at x=2 and -2. Can you explain why the function and these two tangent lines are symmetric?


    2. What is the approx. slope of the tangent to the curve $\displaystyle x^3+y^3=xy$ at x=1
    I got dy/dx for this problem...but i don't know how to get y when i plug in 1

    Thank you
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  2. #2
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    Quote Originally Posted by Linnus View Post
    1. f(x)= 2^x^2
    Find the point of the intersection of the tangent lines to f(x) at x=2 and -2. Can you explain why the function and these two tangent lines are symmetric?


    2. What is the approx. slope of the tangent to the curve $\displaystyle x^3+y^3=xy$ at x=1
    I got dy/dx for this problem...but i don't know how to get y when i plug in 1

    Thank you
    Hello,

    to 1.):

    To calculate the equation of the tangent lines you need a) the coordinates of the tangent point and b) the gradient of the graph of the function at the tangent point.

    $\displaystyle f(x) = 2^{x^2}~\implies~f(2)=2^{2^2} = 16$. Thus the tangent point is $\displaystyle T_1(2, 16)$ and $\displaystyle T_2(-2, 16)$

    To calculate the gradient (=first derivative) I rearrange the term of the function to:

    $\displaystyle f(x) = \left(e^{\ln(2)}\right)^{x^2}=e^{x^2 \cdot \ln(2)}~\implies~f'(x)=2x\cdot \ln(2) \cdot 2^{x^2}$. Use chain rule!

    Therefore the slope of the tangent line is $\displaystyle m_1=f'(2)=64 \cdot \ln(2)$ or $\displaystyle m_2=f'(-2)=-64 \cdot \ln(2)$

    Use the coordinates of the tangent point and the value of the slope to calculate the equation of the tangentline.

    $\displaystyle t_1:y=64 \cdot \ln(2)\cdot x \underbrace{- 128 \cdot \ln(2) + 16}_{\text{-72.7228...}}$ and

    $\displaystyle t_2:y=-64 \cdot \ln(2)\cdot x \underbrace{- 128 \cdot \ln(2) + 16}_{\text{-72.7228...}}$

    Calculate the intersection between these 2 lines and you'll get $\displaystyle C(0, - 128 \cdot \ln(2) + 16)$

    Two graphs f and g are symmetric to the y-axis if their equations satisfy the condition
    $\displaystyle f(x) = g(-x)$ for every x. You can easily show that the equation $\displaystyle t_2$ follows from $\displaystyle t_1$ if you change the x into (-x).

    to 2.:

    I can't help you to solve y - ył = 1

    Maybe you have to use Newton-Raphson method to get an approximate value for y.
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