# how to find inverse

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Sabri Kenno on 16 Oct 2019
Answered: Walter Roberson on 16 Oct 2019
hi. I have a question. how can I solve this: if f(x)=x^2/1+sqrt x how can i find finvese(2)?

darova on 16 Oct 2019
What have you tried?
Dimitris Kalogiros on 16 Oct 2019
Which exactly is your function f(x) ?
Is it the following ?
f(x)=(x^2) / ( 1+sqrt(x) )
And are you trying to solve the equation y=f_inverse(2) ?

Sabri Kenno on 16 Oct 2019
yes its followig:
If f(x)=x^2/(1+sgrtx) find finverse(2) corect to 5 decimal places

Dimitris Kalogiros on 16 Oct 2019
Edited: Dimitris Kalogiros on 16 Oct 2019
Generally, you must have in mind that :
Taking that under consideration you can use the following piece of code
clear; clc;
syms x
% definition of f(x)
f(x)=(x^2) / (1+sqrt(x))
% f(x) is an increasing function of x
fplot(f(x), [0, 3]);
grid on; xlabel('x'); ylabel('f(x)')
% find x0, where f(x0)=2
x0=vpasolve(f(x)==2)
% x0 is the wanted value, since 2=f(x0) <=> finv(2)=x0
% === verification of our solution ===
%find inverse function
finv(x) = finverse(f);
% calculation of finv(2)
x0_verification=vpa(finv(2))
If you don't aim to use symbolic toolbox, you can use simple matlab instructions in order to calculate value x0:
clear; clc;
f=@(x) (x^2)/(1+sqrt(x));
y0=2;
%initial guess for finverse(y0)
x0=0;
%initialization for iterations
yerror=f(x0)-y0;
yerror_previous=yerror;
dx=0.1;
%loop for finding x0
while abs(yerror)>1E-5
% we have cross over y0, step must become less
if yerror*yerror_previous<0
dx=0.1*dx;
end
%store current error for next iteration
yerror_previous=yerror;
x0=x0-sign(yerror)*dx;
yerror=f(x0)-y0;
end
fprintf(' finverse(2)= x0 = %f \n', x0);

Walter Roberson on 16 Oct 2019
target_value = 2;
inverse_for_target = fzero(@(x) f(x)-target_value, initial_guess);