@vv  Thank you for this improvement.

@vv  Maple copes with this example, if a little more help it (in  f1  we manually split  f  into 2 summands):

restart;
f := (3*(x+exp(x))^(2/3)+(x^2+3*x)*exp(x)+4*x^2)/((x+exp(x))^(2/3)*x);
f1:=3/x+``(((x^2+3*x)*exp(x)+4*x^2)/((x+exp(x))^(2/3)*x));
IntegrationTools:-Expand(int(f1, x));
normal(diff(%, x))  # Checking

@magiblot  From the plot it is clear that the maximum point lies in the third quadrant (x<0, y<0). Then by the mouse I just turned the plot (top view) and took a close point, lying in a triangle (x=-5, y=-5) as an initial point.  For  the minimum point everything is similar.

@Rock  Execute  restart  command at first.

@kuwait1 

KK := unapply(diff(n*x^2, x), x, n); 
KK(2, 2);

@jacob123 

Download calc.mw

@mqb  See the edited version above.

@vv  OK, probably you are right. If we give the same initial conditions for a slightly larger t (i took t = 0.1), then a solution exists, in this case f''(0.1) is very large:

l := t -> 0.5*tanh(0.5*t):
deq := diff(f(t), t)*l(t)*(diff(f(t), t, t)*l(t)+9.8*sin(f(t)))+diff(l(t), t)*(diff(f(t), t)^2*l(t)-9.8*cos(f(t))+4*(l(t)-0.5)) = 0, diff(f(t), t, t)=g(t):
sol := dsolve({deq, f(0.1) = 0, D(f)(0.1) = 0.1}, {f(t),g(t)}, numeric);
plots:-odeplot(sol,[t,f(t)],t=0.1..1);
eval(g(t),sol(0.1));
 

@Carl Love   It is better to take the exponent at x of 1.5 instead of 2:

plot(`if`(x>=0 and x<=0.982, 0.015*x^1.5*sin(280*x), NULL), x=0..1.2, -0.02..0.02, color=blue, size=[800,400], gridlines);

I saw two absolutely identical questions for the last 1-2 days and removed the last one as spam. It seems that the second question was automatically deleted. Probably I accidentally pressed the wrong button. I ask you to excuse me for this imprudence.

@John Fredsted 

getMinMax(53^3);

                                                3, 53

Should be   53, 53

See the thread with the same question  http://www.mapleprimes.com/questions/202504-Strange-Behavior-Of--Plotscontourplot

@gtbastos   Try:

restart;
M:=combinat:-permute([0$16,1$16],16):
S:=[seq(`if`(LinearAlgebra:-Modular:-Rank(3, Matrix(4,m, datatype= float[8]))>=3, m, NULL), m=M)]:  # The list of all such matrices, represented as the lists formed of the rows
nops(S);   # The total number of such matrices

@vv  For  x = 0.05 .. 2, y = 0 .. 2  the range of the function is bounded, but rather large. If we increase the lower bound for x-variable, then everything is OK even for the standard grid:

plots:-contourplot(1/(x^2+y^2), x = 0.5 .. 2, y = 0 .. 2, contours = 5, scaling = constrained);

@brian bovril 

seq(max(seq(add(d[s[i]]*x[s[i],s[i+1]], i=1..nops(s)-1) , s=t))<=K, t=Tour2);

d[1]*x[1, 2]+d[2]*x[2, 3]+d[3]*x[3, 4] <= K, max(d[1]*x[1, 2], d[1]*x[1, 3]+d[3]*x[3, 4]) <= K, max(d[1]*x[1, 3], d[1]*x[1, 2]+d[2]*x[2, 4]) <= K, max(d[1]*x[1, 4], d[1]*x[1, 2]+d[2]*x[2, 3]) <= K, max(d[1]*x[1, 2], d[1]*x[1, 3], d[1]*x[1, 4]) <= K