Intersection of tangent lines

**Linnus** · October 21st 2007, 10:04 PM

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 $x^3+y^3=xy$ 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

**earboth** · October 21st 2007, 11:26 PM

Originally Posted by Linnus

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 $x^3+y^3=xy$ 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.

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

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

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

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

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

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

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

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

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

to 2.:

I can't help you to solve y - y³ = 1

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

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