# Function Intersection using Newton's Method

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SB on 19 Nov 2012
Edited: Torsten on 27 Nov 2022
Hi everyone, I'm trying to write a function that finds a point at
which two functions f(x) and g(x) intersect, i.e., f(x) = g(x). I'm using
Newton's Method and making and h(x)=f(x)-g(x) and h'(x) as well, but I'm not getting the right x-value. Please help me debug my code!
% function x = fgIntersect(f, df, g, dg, x0, tol, maxIter)
h=f(x0)-g(x0)
dh=df(x0)-dg(x0)
k=1;
while k<=maxIter
x=x0-h/dh;
if abs(x-x0)<tol*abs(x)
return
end
x0=x;
k=k+1;
end
end
##### 2 CommentsShowHide 1 older comment
SB on 19 Nov 2012
Edited: SB on 19 Nov 2012
format compact; format long;
f = @(x) exp(x) - 3;
df = @(x) exp(x);
g = @(x) sqrt(x);
dg = @(x) .5*x^(-.5);
x = fgIntersect(f, df, g, dg, 1, 1e-6, 50)
x should equal 1.434542442506692
Another case:
format compact; format long;
p1 = [1 -2 3 -8];
p2 = [1 -3 2 -4];
f = @(x) polyval(p1,x);
df = @(x) polyval(polyder(p1),x);
g = @(x) polyval(p2,x);
dg = @(x) polyval(polyder(p2),x);
x = fgIntersect(f, df, g, dg, 2, 1e-6, 50)
x should be 1.561552842846145

Matt J on 19 Nov 2012
You're not updating h and dh within your loop.
Torsten on 26 Nov 2022
Edited: Torsten on 27 Nov 2022
f = @(x) exp(x) - 3;
df = @(x) exp(x);
g = @(x) sqrt(x);
dg = @(x) .5*x^(-.5);
x = fgIntersect(f, df, g, dg, 1, 1e-6, 50)
x = 1.4345
p1 = [1 -2 3 -8];
p2 = [1 -3 2 -4];
f = @(x) polyval(p1,x);
df = @(x) polyval(polyder(p1),x);
g = @(x) polyval(p2,x);
dg = @(x) polyval(polyder(p2),x);
x = fgIntersect(f, df, g, dg, 2, 1e-6, 50)
x = 1.5616
function x = fgIntersect(f, df, g, dg, x0, tol, maxIter)
h=f(x0)-g(x0);
dh=df(x0)-dg(x0);
k=1;
while k<=maxIter
x=x0-h/dh;
if abs(x-x0)<tol*abs(x)
return
end
x0=x;
h = f(x0)-g(x0);
dh = df(x0)-dg(x0);
k=k+1;
end
end