Simple question: How to find the 'x' at a certain value of y(x) equation?
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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 Comments
Dyuman Joshi
on 11 Oct 2023
Moved: Sam Chak
on 11 Oct 2023
Did you tried the approach that is mentioned in the accepted answer?
Accepted Answer
Star Strider
on 28 Feb 2016
This works:
x_for_y30 = fzero(@(x)y(x)-30, 50)
x_for_y30 =
14.0341
6 Comments
More Answers (1)
John BG
on 28 Feb 2016
Edited: John BG
on 29 Feb 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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