We can switch the order of the interpolation to solve this problem. The problem appears regular as measured by the gradient. func = interp1(t,f,time) ДЇsolve completed because the vector of function values is near zeroДЄs measured by the default value of the function tolerance, and Since we use interpolation here, we will get an approximate answer. The second function will look like 0 = 6.25 - f(t).

The second function will be equal to zero at the time. Then, we setup another function that we can use fsolve on. We setup an equation that interpolates f for us. What if we want to know the time that f=0.2? We can approach this a few ways. It is easy to interpolate a new value of f given a value of t. That is because you are trying to estimate the value of an exponential function with a cubic polynomial, it fits better than a line, but it can't accurately reproduce the exponential function. For this example, it only slightly improves the interpolated answer. We can tell Matlab to use a better interpolation scheme: cubic polynomial splines like this. The linearly interpolated example isn't too accurate. The function we sample above is actually f(t) = exp(-t).

For example, suppose we have this data that describes the value of f at time t. We often have some data that we have obtained in the lab, and we want to solve some problem using the data.

interpolate on f(t) then invert the interpolated number.

method 2: switch the interpolation order.

Matlab interp1