How do I find x for a given value of y?

I'm stuck on this, and no one seems to understand the question. I've defined several variables and made x an array of values. I then made an anonymous function and used fplot to graph it's outputs(F in this case) for every x value. Now, I just need to find the x value that give me F=90. The only way I can think to do it is to solve for x by hand and then type that into matlab but there has to be a simpler way. Any body know how to do this? Here's up to where I'm stuck:

 % Clear all windows and variables 
 clc 

 % Input the values for mass, height, friction coefficient, and gravity 
 m=18; % Mass (kilograms) 
 h=10; % Height (meters) 
 mu=0.55; % Friction coefficient (no units) 
 g=9.81; % Gravitational acceleration (m/s^2) 

 % Input a reasonable range for x 
 x=[-100:100]; 

 % Calculate F given the range of x 
 F=@(x) (mu.*m.*g.*(h.^2 + x.^2).^(1/2))/(x+mu.*h); 

 % Plot F as a function of x 
 fplot(F,[0,100]) 
 title('Force versus Distance') 
 xlabel('Distance (meters)') 
 ylabel('Force (newtons)') 
 grid on 

 % Find the x value that gives F=90 
 ??????????????

[Lucijs Milicevic]

Hello, I also have question on this topic.

I need to calculate more than one x value, I have an array of function values for which I need to calculate the x. And my function is cumulative distribution function CDF. I would like iyou can reply me.

CDF = (1/2)*(1+erf(p-mi/(sigma*sqrt(2))));

for example: p=2:0.1:6, mi=4, sigma=0.7...

And function values are as matrix 5x1: 0.78

0.098

0.29

0.47

0.98

[Walter Roberson]

(1/2)*(1+erf(p-mi./(sigma*sqrt(2))))

[Brian Wilkinson]

Here is a code I created that solves for x given y.

% This code solves for x given a value of y. Just set the value for x_low,
% x_high, and y. Enter the formula in terms of x inside the for loop and
% set it equal to y_i. 
x_low=-100; %the lowest value of x that the for loop could possibly output
x_high=100; %the highest value of x that the for loop could possibly output
y=90; %known value of y
x=(x_low+x_high)/2;
for i=1:1000
 m=18; % Mass (kilograms) 
 h=10; % Height (meters) 
 mu=0.55; % Friction coefficient (no units) 
 g=9.81; % Gravitational acceleration (m/s^2) 
 
 y_i=(mu.*m.*g.*(h.^2 + x.^2).^(1/2))/(x+mu.*h);
    
    if y_i<y %the greater than or less than sign may need to be flipped to get a correct value of x after the for loop completes
        x=x-((x_high-x_low)/(2^(i+1))); 
    else
        x=x+((x_high-x_low)/(2^(i+1))); 
    end
end
fprintf('x= %f',x)

Answer 1

[Image Analyst]

Add this code to the bottom of your code:

Fx = (mu.*m.*g.*(h.^2 + x.^2).^(1/2)) ./ (x+mu.*h);
% Plot the function.
figure;
plot(x, Fx, 'b-');
grid on
% Find the x value that gives F=90.
% This happens when the difference between Fx and 90 is a minimum.
[difference, index_At_F_Equals_90] = min(abs(Fx-90))
% Get the x value at that index.  Print to command window.
x90 = x(index_At_F_Equals_90)

[Dan Teep]

Awesome, thanks a ton! I cut out the plot parts but the rest did the job.

[Spottswood]

How do I use this if instead of 90 I'm looking for something like (-3.5). Everytime I just try to switch it out, it gives me an error saying "subscript indices must be integers and non-negative" or close to that. Please help!

[Walter Roberson]

target = -3.5;
[difference, index_At_F_Equals_target] = min(abs(Fx-target));
xtarget = x(index_At_F_Equals_target);

[Spottswood]

!! Thank you so much! You're a lifesaver

[Kalyan Dash]

Hi Walter Your code works perfect. But can you please elaborate the code. How it is calculating the answer and what values are stored in the variables named difference and index_At_F_Equals_target. And how to use xtarget function.

[Walter Roberson]

abs(Fx-target) will be closer to 0 at the place where the values in Fx are closer to the target value. The location in Fx that is closest to the target will have the smallest absolute difference compared to the target. min() applied to that vector determines the location where the minimum value is. min() with a single output would be just the minimum value itsef; min() with two outputs like is given here also returns the index in the vector at which the minimum was found. So after the min() line, the variable difference will reflect how close you were able to get to the target, and index_At_F_Equals_target will be the location (index) in Fx that was closest. Then you access the vector of x values at that index to determine which x value it was that gave rise to the Fx that was closest to the target. xtarget will not be function after this: it will be the numeric x value at which Fx became closest to the target.

[Kalyan Dash]

Thanks Walter. It is much clear now after reading the description. I have used it in my code and it worked.

Answer 2

[Walter Roberson]

fzero((x) F(x)-90, [0, 100])

[Dan Teep]

Hmmm, I'm getting an error message at the column with the function name. I guess I'll try using a function file real quick instead of an anonymous one.

[Walter Roberson]

Darn, somehow I deleted a character

fzero(@(x) F(x)-90, [0, 100])

Answer 3

[Deep Patel]

how to solve 4*exp(-x^2)*sin(x) = 1 for x in matlab?

[Stephen23]

>> fun = @(x)4*exp(-x^2)*sin(x)-1;
>> x = fzero(fun,1)
x =  0.27266

[Deep Patel]

Thank you for answering..But how can I do this in mupad?

[Walter Roberson]

For the symbolic toolbox you would use

syms x
solve(4*exp(-x^2)*sin(x) == 1)

If you are trying to work in a MuPad notebook (which are probably going to be completely gone by R2018b), then

numeric::solve(4*exp(-x^2)*sin(x) = 1)

[Deep Patel]

Thank you very much