Use Newton's Method to approximate, to three decimal places,...

Use Newton's Method to approximate, to three decimal places, the $x$ -coordinate of the point of intersection of the graphs of the two equations. Use a graphing utility to verify your result.
$$
y=\ln x, \quad y=-x
$$

Key Concepts

Intersection of Functions

Finding the intersection of two graphs involves equating the two function expressions and solving for the variable. This transforms the problem into finding the roots of a new function defined as the difference between the two original functions. The solutions to this equation represent the x-coordinates where the two graphs intersect.

Graphical Verification

Graphical verification involves using a graphing utility or software to visualize the functions and confirm that the computed intersection point is accurate. This serves as a complementary check to analytical or numerical methods, ensuring that the approximated solution aligns with the actual behavior of the functions.

Newton's Method

Newton's Method is an iterative numerical technique used to approximate the roots of a function. It involves using the formula x??? = x? - f(x?)/f'(x?) to generate successively better estimates of a root, provided that the function is differentiable near the root and the initial guess is sufficiently close to the actual root.

Transcript

00:01 Okay, we want to use newton's method to find where y equals lnx and y equals minus x intersect.

00:07 So the first thing we have to do is make that into a function.

00:11 So we're trying to solve lnx equals minus x.

00:16 So our function will be natural log of x plus x.

00:24 Then we need its derivative, which will be 1 over x plus 1 or 1 plus x over.

00:32 X.

00:35 Okay, and then we need a point to start with because we're going to use this formula up here where you have a starting point, x, 0, and then you find f of x of 0 and the derivative of x of 0, and then subtract, and that will give you the next x to try.

00:55 Okay, so i'm going to draw a picture of lnx and minus x and look to see where they intersect.

01:02 1, 2, 3.

01:04 So lnx goes through e1, 10, 1 over e minus 1.

01:11 So there's lnx, and then y equals minus x.

01:17 So i'm going to guess 1 for my starter.

01:22 So when i plug into f of x, i get ln of 1 plus 1, which is 1.

01:28 And in the derivative 1 plus 1 over 1 which is 2 so 1 minus 1 half which gives me 1 half so 1 half is my next place so then i need ln of 1 half plus 1 and then i need 1 plus 1 half over a half which is 3 okay but now i need my calculator for the rest of it so i've programmed it in to calculate for me if i let's see so i need in my table okay so i'm plugging in point five oh it's not that point five so for this i get minus point one nine three one four seven so i have one half minus minus point one nine three one nine seven over three and so that gives me point 566438.

02:56 0 .56438...

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