Solving an equation for a variable

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Raymond Elliott
Raymond Elliott 2021 年 2 月 22 日
コメント済み: Walter Roberson 2021 年 2 月 23 日
I am given the following equation,
fx =@(x) ((15*3)/(3*pi^4*70*5.29E-6))*(48*3^3*cos((pi*x)/(2*3))-(48*3^3)+(3*pi^3*3*x^2)-(pi^3*x^3));
If I wanted to find the value of x at say fx = 10, is there a shortcut to do so in MATLAB?

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Walter Roberson
Walter Roberson 2021 年 2 月 22 日
The trick is to create a new function, such as fx10, that is fx(x)-10 and then do root finding (fzero, fsolve, vpasolve) on that new function.
  5 件のコメント
Raymond Elliott
Raymond Elliott 2021 年 2 月 22 日
One last question, how were you able to find that there were two roots that were close to zero? Did you plot the function to see or did you do it some other way?
Walter Roberson
Walter Roberson 2021 年 2 月 23 日
The second input is a starting guess.
I used a negative starting guess and a modest positive starting guess. Afterwards I plotted to confirm.

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