Simple question: How to find the 'x' at a certain value of y(x) equation?
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A
el 28 de Feb. de 2016
Comentada: Star Strider
el 17 de Abr. de 2024 a las 13:16
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!
4 comentarios
Dyuman Joshi
el 11 de Oct. de 2023
Movida: Sam Chak
el 11 de Oct. de 2023
Did you tried the approach that is mentioned in the accepted answer?
Respuesta aceptada
Star Strider
el 28 de Feb. de 2016
This works:
x_for_y30 = fzero(@(x)y(x)-30, 50)
x_for_y30 =
14.0341
8 comentarios
Thanh Thuy Duyen Nguyen
el 17 de Abr. de 2024 a las 12:46
If I have x and want y then how could I compute that?
Star Strider
el 17 de Abr. de 2024 a las 13:16
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
yq = 30;
x_for_y30 = interp1(y(x), x, yq) % Get 'x' For Specific 'y'
y_for_x2pi = interp1(x, y(x), 2*pi) % Get 'y' For Specific 'x'
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))
.
Más respuestas (1)
John BG
el 28 de Feb. de 2016
Editada: John BG
el 29 de Feb. de 2016
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
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