if true clc;clear;close;format longsyms xf=0.0000095862*((x+1.4051)^0.44267)*((6.4051-x)^5.5867);g=diff(f,x);g2=diff(f,2);h=solve(g==0)h2=solve(g2==0)A=subs(f,x,h)end
A must be double? Why it is sym in workplace ? Besides I have some warnings; Warning: Cannot solve symbolically. Returning a numeric approximation instead.
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