how to find inverse

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]

What have you tried?

[Dimitris Kalogiros]

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]

yes its followig:

If f(x)=x^2/(1+sgrtx) find finverse(2) corect to 5 decimal places

[Venkata swapna]

1/invcos(x)*(cos(x)-1) in coding

Answer 1

[Dimitris Kalogiros]

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;
    
    %adjust x0 
    x0=x0-sign(yerror)*dx;
    
    yerror=f(x0)-y0;
    
end

fprintf(' finverse(2)= x0 = %f \n', x0);
        

Answer 2

[Walter Roberson]

target_value = 2;
inverse_for_target = fzero(@(x) f(x)-target_value, initial_guess);