Confuse with function handles, matrix and single point

Hello, I got homework with this a little bit complex function.

I calculate values of the function with input is a vector and single points. Same function but the results are pretty difference!!! (See on the graph.

Anyone can explain it?. Thanks a lot!

Here 's my image and code .

x   =-2:.01:2;
f   =@(x) x.*sqrt(x.^4+1).*log(4-x.^2)+...
    6*sin(4+x.^2)/(1+exp(-x))-...
    7*x;
%----------
grid on
hold on
plot(x,f(x),'*r-')
%------------------
for i=1:length(x)
    plot(x(i),f(x(i)),'o')
end
%---------------------------------------------------
% ff  = x.*sqrt(x.^4+1).*log(4-x.^2)+...
%     6*sin(4+x.^2)/(1+exp(-x))-...
%     7*x;
% plot(x,ff,'g')
% % -----------------------------------------
% for i=1:length(x)
%     xx=x(i);
%     fff  = xx.*sqrt(xx.^4+1).*log(4-xx.^2)+...
%     6*sin(4+xx.^2)/(1+exp(-xx))-...
%     7*xx;   
%     plot(xx,fff,'y*')
% end

Answer 1

[Amit]

In defining f, you didn't do the function right. To take a value of x as a vector, you need to use ./ instead of just / . In think the proper code will be

x   =-2:.01:2;
f   =@(x) x.*sqrt(x.^4+1).*log(4-x.^2)+...
    6*sin(4+x.^2) ./ (1+exp(-x))-...
    7*x;

This will be same for both scaler and vector inputs.

[Bac VU]

Thanks Amit for the helpful question. It works perfectly,

Matlab usually return error "dimension mismatch" when I try to multiply vectors, so I didn't note that Matlab do divides same length vector in this case.