We 1st introduce in this section the function notation as distinct from the expression notation.  you are given the function and wish to operate on it (graph, evaluate, differentiate, ) This is usually advisable to use the function notation which assigns to an independent variable x, or u, or t, a value f(x), or whatever, or Q(u), or g(t). The Consider polynomial P(x)=x^2-5*x-14 and we write it as a function:

> P:=x->x^2-5*x-14;


Now evaluate it at x=1.75.

> P(1.75);


Now find its roots.

> solve(P(x)=0,x);


Expression =0 may be omitted from the solve command in which case it is understood that you are solving P(x)=0. Right hand side is not 0 then make the necessary adjustments.

> solve(P(x)=4,x);


Solve( ) command will only find exact roots. This can't find any it returns no answer or an unusable one. To instance

> solve(P(x)=4*sin(x),x);


Now we write the polynomial as an expression.

> P:=x^2-7*x-11;


> P(1.75);


Evaluation must be done differently and requires more effort.

> subs(x=1.75,P);


Solve command may be used as before.

> solve(P,x);


> evalf(%);


Command evalf( ) uses the very accurate floating point arithmetic. This must be used whenever irrational numbers occur and we recommend it be used at all times. For go directly to decimal approximations of roots use the fsolve( ) command.

> fsolve(P,x);


fsolve( ) will only find real roots, this produces approximate values.

> Q:=x->x^3+5*x^2+6*x+8;


> fsolve(Q(x),x);


> solve(Q(x),x);


> evalf(%);


Solving an equation is the 1 shot deal we can conveniently type in the function in the solve( ) command.

> solve(x^4-x^3-x^2+3*x-6,x);


> evalf(%);


if the equation is not a polynomial then the fsolve( ) command, initiates a numerical 0 finding scheme, must be used. For find the intersection point of the graphs of 2*x and cos(x) we must solve the transcendental equation cos(x)-2*x=0.

> solve(cos(x)-2*x,x);


We must use fsolve( ).

> fsolve(cos(x)-2*x,x);





Using the fsolve( ) command to find real roots some graphing may be needed since the fsolve( ) command is very nearsighted. PROBLEM: Find the 0 of the following function.

> g:=x->x-x^2+x^4/(10*x+4);


> fsolve(g(x),x);


Fsolve( ) command has only found the obvious 0. The little analysis will convince us there are 2 more positive 0 and we use some graphing.

> plot(g(x),x=0..4);


> plot(g(x),x=4..12);


Plots shown we can give ranges for fsolve( ) to find the other 0.

> fsolve(g(x),x=0.8..1.2);


> fsolve(g(x),x=8..10);


We only need rough estimates of the 0 we can box the graphs then put the cursor on the x-axis crossing and read off an approximate value.

PROBLEM: Find the eighth positive zero of sin(x)-x/40. It time we want to graph two functions simultaneously and the easy way is the command plot({f(x), x=a..b), g(x)}.

> plot({sin(x),x/40},x=0..50);


> f:=x->sin(x)-x/40=0;


That we inserted the =zeero term - this is good practice.

> solve(f(x),x);


> fsolve(f(x),x);


Nothing happened except the obvious. The Box graph and use the cursor to estimate the x-coordinate of the intersection point of the two graphs, Give a range for x.

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