Using Newtons Method

[DumbJock]

The question reads "Find the point on the graph of y=e^(-x) where the normal line to the curve will pass through the origin. (Use Newton's Method)"
I have taken y=e^(-x) and have gotten it to e^(x)(x)-1/(e^(x))=0 but do not know what to do next. I am not sure how to use newton's method for this problem either. SOS.

 

[MarkFL]

I would proceed as follows:

\(\displaystyle f(x)=e^{-x}\)

1.) Find the slope \(\displaystyle m\) of the normal line:

\(\displaystyle -\dfrac{1}{m}=\dfrac{d}{dx}(f(x))\)

Observe that the equation of the normal line must be \(\displaystyle y=mx\).

2.) Since the normal line passes through the origin \(\displaystyle (0,0)\) and \(\displaystyle \left(x,e^{-x} \right)\), use these two points to write another expression for \(\displaystyle m\).

3.) Equate the two expressions for \(\displaystyle m\) found in 1.) and 2.), then use Newton's method to approximate the root of the resulting equation.

Use this approximation in the equation for the normal line from 1.).

Let us know what you find!

 

[DrPhil]

The question reads "Find the point on the graph of y=e^(-x) where the normal line to the curve will pass through the origin. (Use Newton's Method)"
I have taken y=e^(-x) and have gotten "it" [what is "it"?] to
e^(x)(x) - 1/(e^(x)) = 0
but do not know what to do next. I am not sure how to use newton's method for this problem either. SOS.

Looks like a good start - let me check that you have the normal line at point (x0, y0).

y(x) = e^{-x},.......dy/dx = - e^{-x}
y0 = e^{-x0},.......[dy/dx]0 = - e^{-x0}

slope of normal = m = -1/[dy/dx]0 = e^{x0}
eq. of normal = n(x) = y0 + m*(x - x0)
..............................= e^{-x0} + e^{x0}*(x - x0)
y-intercept = b = y0 - m*x0 = x^{-x0} - x0*e{x0}

Looks like my "b" is the negative of your "it," so what you have to do next is solve the equation
......f(x) = x*e^x - 1/e^x = 0
For Newton's method, you need the derivative
......f'(x) = . . .
Then select a starting value for x = x0 (maybe ~0.5?), evaluating f(x0) and f'(x0)
and iterate
x{n+1} = x{n} - f(x{n}) / f'(x{n})

Repeat till the error is small enough to suit you.