Simple question: How to find the 'x' at a certain value of y(x) equation?

This may be a simple question. But let's assume I have one ugly equation:

x = [0:10];
y = @(x) x.^2.*12./23./23.9.*log(x).^2

How do I find the value of 'x' where y = 30?

Thanks!

[Ishank Gupta]

I am new to matlab and in general coding.

I was writing a code for taylors series in matlab so I thought what if i input the function from user itself and the solve for that. But to do that i need to find the function value at x=0, How should i proceed with this.(just curious)

Thanks a lot..

[Ceylin]

How to find the value of x in this equation: sin(x)=0.2

[Dyuman Joshi]

Did you tried the approach that is mentioned in the accepted answer?

[Sam Chak]

@Ceylin, Which intersection do you want to solve for x?

syms f(x)
f(x) = sin(x);
fplot(f, [-2*pi, 2*pi]), grid on    % draw left  side of Eqn
yline(0.2, 'r-')                    % draw right side of Eqn
xlabel('x')
legend('sin(x)', '0.2')

Answer 1

[Star Strider]

This works:

x_for_y30 = fzero(@(x)y(x)-30, 50)

x_for_y30 =
    14.0341

[A]

Thanks! What does the 50 mean in that formula?

[Star Strider]

My pleasure!

The 50 is a starting guess for fzero. It needs to have an initial estimate of the solution (as do all iterative gradient-descent algorithms, such as nonlinear parameter estimation or function minimisation algorithms such as fminsearch). The fzero function will then start there to search for a zero-crossing. Since I subtracted 30 from your function to create a zero-crossing there, it then uses the initial estimate to find the value of ‘x’ at the zero-crossing, in this instance, the value of your function at y=30 (that it thinks is y=0).

[A]

So does that number need to be 'less' than the anticipated result?

Thanks

[Star Strider]

No. The number, 30 here, can be anything the function is defined at for any value of ‘x’. So for instance:

y = @(x) sin(x);

would return NaN because sin(x) would never equal 30.

If you’re interested in the initial estimate for fzero, that can be anything, provided that (1) the function is defined at the initial estimate, (2) is defined for all values between it and the value of ‘x’ at the zero-crossing, and (3) that a zero-crossing exists. So there are some restrictions, but only those imposed by the continuity (and differentiability) of the function, and the behaviour of the function on the interval of interest. For instance, if you have a function with more than one zero, in order to find all of them, you have to loop through appropriate initial estimates for the zero-crossings. That can get complicated, so I won’t go into it here.

[A]

Thanks! That was a great explanation!

[Star Strider]

As always, my pleasure!

That’s what we’re here for!

[Thanh Thuy Duyen Nguyen]

If I have x and want y then how could I compute that?

[Star Strider]

Actually there is an additional way of solving this, using interp1, especially if the actual function is not available —

x = [0:15]+eps;
y = @(x) x.^2.*12./23./23.9.*log(x).^2

y = function_handle with value:
    @(x)x.^2.*12./23./23.9.*log(x).^2

yq = 30;

x_for_y30 = interp1(y(x), x, yq)                        % Get 'x' For Specific 'y'

x_for_y30 = 14.0322

y_for_x2pi = interp1(x, y(x), 2*pi)                     % Get 'y' For Specific 'x'

y_for_x2pi = 2.9555

figure
plot(x, y(x))
grid
hold on
plot(x_for_y30, 30, 'ms', 'MarkerFaceColor','m')
plot(2*pi, y_for_x2pi, 'cs', 'MarkerFaceColor','c')
hold off
xline(2*pi,'--k', '2\pi')
yline(y_for_x2pi, '--k', sprintf('y=%.2f  for  x=2\\pi',y_for_x2pi))
xline(x_for_y30, '--k', sprintf('x=%.2f  for  y=30',x_for_y30))

.

Answer 2

[John BG]

Alpha

If you plot the following

    x=[-100:.1:100] 
    f = @(x) x.^2.*12./23./23.9.*log(x).^2
    y=f(x)
    plot(x,y)
    grid on

place the marker on the point that shows y=30 f(x) is not symmetric, it has 2 zeros, and f=30 on 2 places:

x01=14.04 
x02=-29.5

if what you really mean is:

f2 = @(x) x.^2.*12./(23.*23.9).*log(abs(x)).^2

then the function is symmetric and there are 2 values of x that satisfy your question:

x01=14.04 
x02=-14.04

Compare both functions and y=30

If you find this answer of any help solving this question, please click on the thumbs-up vote link

thanks in advance

John

[e-mail]