How to find second intersection point?

Okay, so

My task is to find the intersections points of functions f1 = 1/x and f2 = sqrt(5./2 - (x^2))

I've found one of the intersections points using:

Intersections=find(abs(f1-f2)<=(0.05));

xvalues=x1(Intersections);

But looking at the graphs I see there are two intersections points, so how do I find the other?

Answer 1

[Star Strider]

It is easiest to do this symbolically —

syms x 
f1 = 1/x;
f2 = sqrt(5./2 - (x^2));
Intx = solve(f1 == f2)

Intx = 

Intxd = double(Intx)

Intxd = 2×1

1.4142 0.7071

.

[Amanda]

Is there a way to solve this without using syms? We're not allowed to use it.

[Star Strider]

Yes, although this is slightly complicated by the fact that ‘f2’ is complex —

x = linspace(0, 5);
f1 = 1./x;
f2 = sqrt(5/2 - (x.^2));

figure
plot(x, real(f1),   x, real(f2))
grid

idx = find(diff(sign(real(f1)-real(f2))));

for k = 1:numel(idx)
    idxrng = max(1,idx(k)-1 : min(numel(x),idx(k)+1));
    Intx(k) = interp1(real(f1(idxrng))-real(f2(idxrng)),x(idxrng),0);
end

Intx

Intx = 1×2

0.7071 1.4142

EDIT — (17 Sep 2022 at 14:18)

Using fzero —

f1fcn = @(x) 1./x;
f2fcn = @(x) sqrt(5/2 - (x.^2));

for k = 1:2
    Intx(k) = fzero(@(x)real(f1fcn(x))-real(f2fcn(x)), k);
end

Intx

Intx = 1×2

0.7071 1.4142

.

[Amanda]

Thank you very much! Could you explain your foor-loop (not the one in the edit)?

[Star Strider]

My pleasure!

Sure! The fzero function is a root-finding algorithm, and can only return one root at a time, so to find more roots, it is necessary to iterate calls to it with different initial estimates. It will return the closest root to each estimate. Here, one root is closer to 1 and the second root is closer to 2, so I just used the loop counters for the initial estimates here. Usually, more elaborate initial estimates are required. I generally prefer the interp1 approach if it is applicable and if I know the region-of-interest, since it is easier to discover multiple roots or intersections with it, using the ‘idx’ approach to finding them.

[Amanda]

I tried using your code, and got the same values of x. However, looking at the graph, the second x-value is incorrect. We get x = 1.4142, while it seems to be more 1.387

[Star Strider]

The symbolic approach solves for and , agreeing with the numeric approach, so I do not see how any other values could be correct.

Answer 2

[Torsten]

x1 = 0:0.0001:1.5;
f1 = 1./x1;
f2 = sqrt(5./2 - (x1.^2));
Intersections = find(abs(f1-f2)<0.00005);
xvalues=x1(Intersections)

xvalues = 1×2

0.7071 1.4142