Hi,
many options to do this - here you have 3 of them. All options start with your known code and want to know the x-value for y = 35.0048:
x=[0 1 2 3 4 5];
y=[0 20 60 68 77 110];
coeffs = polyfit(x,y,1);
y_val = 35.0048;
coeffs_new = coeffs;
coeffs_new(2) = coeffs_new(2) - y_val;
result1 = roots(coeffs_new);
leads to:
result1 =
1.5000
syms f(x) f(y)
f(x) = coeffs(1) * x + coeffs(2)
f(y) = finverse(f)
which gives:
f(x) =
(729*x)/35 + 79/21
f(y) =
(35*x)/729 - 395/2187
To calculate values with this you will need a function handle:
f_y = matlabFunction(f(y))
which is:
f_y =
function_handle with value:
@(x)x.*(3.5e1./7.29e2)-1.80612711476909e-1
Then you can calculate the x-value belonging to 35.0048:
>> result2 = f_y(y_val)
result2 =
1.5000
result3 = fsolve(@(x)coeffs(1)*x+coeffs(2)-y_val,0)
which results in:
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
result3 =
1.5000
Best regards
Stephan
