Root of exponential function not complete

I am trying to find the roots of the equation
exp(-0.5*x)-0.4/(1+0.5*x)
and with the following code, only one of the 2 roots is found:
syms x
eq1=exp(-0.5*x)-0.4/(1+0.5*x)==0;
sol=solve(eq1,x);
vpa(sol,6)
ans =
-1.64929
But there is another root of this equation which is found using the fzero command
eq1=@(x) exp(-0.5*x)-0.4/(1+0.5*x);
sol=fzero(eq1,[-10 10]);
vpa(sol,6)
ans =4.04463
Again the given solution is only one of the 2. How can I get both of them at any time with one command? Thanks

Antworten (1)

madhan ravi
madhan ravi am 28 Okt. 2018
Bearbeitet: madhan ravi am 28 Okt. 2018

0 Stimmen

syms x
eq1=exp(-0.5*x)-0.4/(1+0.5*x)==0;
fplot(exp(-0.5*x)-0.4/(1+0.5*x)) %always compare the result with the graph
solution1=vpasolve(eq1,x,[-2 0])
solution2=vpasolve(eq1,x,[0 5])
grid on

7 Kommentare

Thanks for answering, but vpasolve also gives only one solution, the same as fzero (4.0446). The -1.64929 is not found
madhan ravi
madhan ravi am 28 Okt. 2018
Did you see the graph it appears there’s only one solution right?
Needless Needless Also
Needless Needless Also am 28 Okt. 2018
Bearbeitet: Needless Needless Also am 28 Okt. 2018
Sorry, but I see two roots. Am I wrong?
madhan ravi
madhan ravi am 28 Okt. 2018
Ah i missed it
madhan ravi
madhan ravi am 28 Okt. 2018
See edited specify the intervals
This works but, what about any other equation? Shouldn't it be a pretty simple task for MatLab to find the roots just with vpasolve(eq1,x)? Of course I can always check the graph but I think you get what I mean.
madhan ravi
madhan ravi am 28 Okt. 2018
Yes I know even I have thought about it ;) but yet no discernible idea in my mind though

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R2018a

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am 28 Okt. 2018

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