In older versions of Maple (for example in Maple 12)  ListTools:-FindMaximalElement  command does not work. Workaround is   ListTools:-Search  or   ListTools:-SearchAll  commands:

L := [1,2,3,7,6,5,4]:

m:=max(L);

pos:=ListTools:-Search(m,L);

                        m := 7

                       pos := 4

limit(n^(1/n), n=infinity);

limit(exp(n)/n^4, n=infinity);

                               1

                           infinity

For visual purposes only.

Example:

restart;

applyrule(-cos(x::anything)=cos(x+pi), sin(t)-cos(t));

subs(pi=Pi, %);

%;

For a correct perception  add the option  view=[-5..0, 0..1]  into Carl's code after  axes=boxed :

For  x>0  this equation is equivalent to the equation  add(ln(x+j), j=0..2015)=0 , so we have:

Digits:=20:

fsolve(add(ln(x+j), j=0..2015), x=0..infinity);

              0.86678004277349545751*10^(-5785)

It is obvious that this is the only positive root.

restart;

sys:=diff(fi1(t),t,t)=(m0-m)/5, diff(fi2(t),t,t)=m/50:

m0:=200:  m:=(fi1(t)-fi2(t))*32.2+(diff(fi1(t),t)-diff(fi2(t),t))*10:

dsol2:=dsolve({sys,fi1(0)=0,fi2(0)=0,D(fi1)(0)=0,D(fi2)(0)=0}):

fi1:=unapply(rhs(dsol2[1]), t): fi2:=unapply(rhs(dsol2[2]), t):

t1:=fsolve(D(fi1)(t)=10);  t2:=fsolve(D(fi2)(t)=10);

plot([piecewise(t>=0 and t<=t1, D(fi1)(t), 10), piecewise(t>=0 and t<=t2

, D(fi2)(t), 10)], t=0..10, color=[red,green], thickness=2); 

These are the very simple problems. See help on Maple commands:

1.  dsolve

2.  plots[fieldplot]   and   plots[fieldplot3d]

3.  plot3d  - 3rd variant in Calling Sequence

I recorded your data as 2 matrices consisting of lists simultaneously correcting several errors:

A:=<[0.55,0.67,0.78,0.89],[0.7,0.8,0.8,0.9],[0.767,0.867,0.93,0.967],[0.72,0.83,0.83,0.93];    

[0.67,0.78,0.89,0.97],[0.8,0.9,1,1],[0.73,0.83,0.867,0.93],[0.66,0.76,0.79,0.90];

[0.78,0.89,0.89,1],[0.8,0.9,1,1],[7.67,8.67,9.3,9.67],[0.55,0.66,0.69,0.79];

[0.78,0.89,0.89,1],[0.06,0.13,0.167,0.267],[0.8,0.9,1,1],[0.76,0.86,0.90,0.97];

[0.78,0.89,0.89,1],[0.8,0.9,1,1],[0.06,0.13,0.167,0.267],[0.76,0.86,0.90,0.97];

[0.74,0.85,0.93,1],[0.67,0.767,0.83,0.9],[0.73,0.83,0.867,0.93],[0.62,0.72,0.83,0.90];

[0.59,0.70,0.74,0.85],[0.567,0.667,0.73,0.83],[0.667,0.767,0.83,0.9],[0.69,0.79,0.86,0.93];

[0.74,0.85,0.93,1],[0.7,0.8,0.9,0.93],[0.567,0.667,0.73,0.83],[0.79,0.90,0.97,1];

[0.70,0.81,0.85,0.96],[0.7,0.8,0.9,0.93],[0.7,0.8,0.8,0.9],[0.59,69,0.76,0.86]>:

V:=<[0.7,0.8,0.8,0.9],[0.8,0.9,1,1],[0.767,0.867,0.93,0.967],[0.43,0.53,0.567,0.667],

[0.73,0.83,0.867,0.93],[0.8,0.9,1,1],[0.067,0.1,0.2,0.3],[0.73,0.83,0.867,0.93],[0.53,0.63,0.667,0.767]>:

Matrix(9,4, (i,j)->zip((a,v)->a*v, A[i,j], V[i,1]));  # Final answer

Example with 4 points:

r:=[12, 56, 29, 78]:

v:=[15, 45, 75, 102]:

map(t->[t[1]*cos(t[2]),t[1]*sin(t[2])], zip((x,y)->[x,y],r,v));

     [[12*cos(15), 12*sin(15)], [56*cos(45), 56*sin(45)], [29*cos(75), 29*sin(75)], [78*cos(102), 78*sin(102)]]

y  is a complex expression, so you can plot only  Re(y)  and  Im(y)  separately or  together as parametric function:

restart;

y := A*(1/x+x*exp(-2*sqrt(-1)*b))+4*sin(h)^2*(2*exp(-sqrt(-1)*b)-3*sin(h)^2*x^(-sin(h))*exp(sqrt(-1)*b*(-sin(h)-1))+3*x^(-sin(h))*exp(sqrt(-1)*b*(-sin(h)-1)))/(3*(1-r))-exp(-2*sqrt(-1)*b)/x-x:

A := (1+r)/(1-r):  r := (1/3)*sin(h)^2:  b := m*Pi*h:  m := 1:  h := 0.05:

plot([Re(y), Im(y)], x=0..3, -1..2, color=[red,blue], thickness=2);

plots[complexplot](y, x=0..3, view=[-1..1,-1..1], thickness=2);

The second plot shows that Rе(y)  and Im(y)  probably  are linearly related.

eq := (4*a^3*b)^(1/2)/(-(a/(4*b))^(1/2))+(4*a^3*b*(4*b/a))^(1/2) = 0:

simplify(eq)  assuming a::positive, b::positive;

simplify(eq)  assuming a::negative, b::negative;

Unfortunately Maple cannot itself find restrictions under which the identity is true.

The problem can be solved without using inequalities:

{solve((1/(a-1)-1/(a+1))*(2*a^2-2)/a = 4/sqrt(2), a)}  minus  {-1, 1};

Rouben Rostamian Your example is not an example of  p versus w. It is the example of a parametrically given function  z->(w(z), p(z))  with  z  as a parameter. Here is the example - the plot  w->p(w)  in the range w=50000..150000 :

restart;
Sys := -(diff(p(z), z)) = 1.289*10^10*(.675+77/(Pi*1.2^2))*(1+x(z))*(1/(Pi*1.2^2*p(z))), diff(x(z), z) = (1.134*10^(-8)*Pi)*1.2^2*(1-x(z))*p(z)/(1+x(z)):
Inc := p(0) = 2*10^6, x(0) = 0:
Sol := dsolve({Inc, Sys}, numeric):
Eq := w = 1560*Pi*(9*z-(1/3)*(z-2.7)^3*((1/3)*2.7^3)):
f := t->fsolve(eval(Eq, w = t), z):
w1 := 50000: w2 := 150000:
z1 := min(f(w1), f(w2)): z2 := max(f(w1), f(w2)):
plot(rhs(Eq), z = 0 .. 7, -100000 .. 200000, labels = [z, w], labelfont = [TIMES, ROMAN, 14]);  # The plot of the function  z->w(z)
plots[odeplot](Sol, [rhs(Eq), p(z)], z = z1 .. z2, color = blue, labels = [w, p], labelfont = [TIMES, ROMAN, 14]);  # The plot of the function  w->p(w)  in the range  w=50000..150000

Of course, it is easier to solve the equation by hand, as M. Hirnyk  wrote, but if you want to solve with Maple, then there are some difficulties. Because all three terms of the equation are non-negative, then the sum is equal to 0 only if each term is equal to 0. But Maple somehow does not solve the simple system  sys . Therefore it is necessary to solve it step by step:

restart;

eq:=abs(a-b)+sqrt(2*b+c)+``(c^2-c+1/4)=0;

L:=op(lhs(eq));

sys:={L[1]=0,L[2]=0,op(L[3])=0};

solve(sys,{a,b,c});  # Incorrect solution

c0:=solve(op(L[3]))[1];  # First we solve the equation  c^2-c+1/4=0

op(solve(eval(sys,c=c0))),c=c0;  # Final answer

If I understand your question then

eval(1.18971*x^0.38480, x=17);  # Or

f:=x->1.18971*x^0.38480:

f(17);

                     3.539303919

                     3.539303919