You need to understand the mathematics. Lost you will be without that (to misquote someone named Yoda.)
For this polynomial:
y=85.*x- 91.*(x.^2)+ 44.*(x.^3)- 8.*(x.^4)+(x.^5)-26;
There seems to be a root near zero. Thus a root is where the function has a value of y=0. But when it is flat like that, this is often an indication there may be multiple roots, or at least several roots that are very close to each other. Or you may have a complex root relatively close to that point too. And the scale on the y axis makes it difficult to be sure.
roots(p)
ans =
2.5907 + 4.4333i
2.5907 - 4.4333i
1.1308 + 0.7012i
1.1308 - 0.7012i
0.5570 + 0.0000i
However, roots tells us there is only one real root of the polynomial. It lies at x = 0.5570. while the other roots are complex.
What happens when you tried to use interp1? What does interp1 do here? Actually, interp1 is a terribly bad idea. When you flip x and y like that, for a function that has a nearly zero slope, then flipping y and x produces a function with nearly infinite slope. And using an INTERPOLATION function to approximate something with a nearly infinite slope is a BAD IDEA.
You used
x = linspace(-100,100,100);
so 100 points, that varied from -100 to 200. That means a spacing of roughly 2 between each point. Do you see how much curvature you have in that function? Then using interp1, which by default uses LINEAR interpolation is just completely misunderstanding what is happening. The only points that are near zero in x at all were:
x(49:53)
ans =
-3.0303 -1.0101 1.0101 3.0303 5.0505
yfun = matlabFunction(y)
yfun =
@(x)x.*8.5e+1-x.^2.*9.1e+1+x.^3.*4.4e+1-x.^4.*8.0+x.^5-2.6e+1
plot(yfun(x(49:53)),x(49:53),'o-')
the problem is, if you use interp1 here, it uses only linear interpolation. And that means it tries to interpolate the connect-the-dots curve drawn there. Using interp1 here to solve for a root of this function is again, just a flat out bad idea. I know that people have told you to do that online. But they were wrong, at least to tell you to just swap x and y, and then to use interp1, if done without any understanding of what is happening.) That is certainly true if you use such a coarsely sampled grid in x.
CAN we do a better job, using interp1 more intelligently? Well, yes. By using a more finely sampled grid, we might do this:
xbetter = linspace(-10,10,100);
xinterproot = interp1(ybetter,xbetter,0)
xbetter = linspace(-10,10,1000);
xinterproot = interp1(ybetter,xbetter,0)
xbetter = linspace(-10,10,10000);
xinterproot = interp1(ybetter,xbetter,0)
So you see that by use of a finely sampled grid, the LINEAR interpolation can do a better job. However, we could also have used a spline interpolant with interp1. This is still a dangerous thing to do with interp1 using a spline, because splines themselves have serious problems.
xbetter = linspace(-10,10,100);
xinterproot = interp1(ybetter,xbetter,0,'spline')
xbetter = linspace(-10,10,1000);
xinterproot = interp1(ybetter,xbetter,0,'spline')
which you can see that even on the moderately coarsely sample grid, did a decent job. I did not need to go to as finely sampled a grid to get 4 digits correct there, when using a spline.
Just use roots, which will produce the correct solution, here accurate to full double precision accuracy. Or, you could have used solve or vpasolve. Since this is a 5th degree polynomial, we cannot compute an algebraic form of the roots of a 5th degree polynomial anyway, so vpasolve is correct here.
vpasolve(y)
ans =
