> f:=x->x^3-3*x+1; > plot(f(x),x=-3..3); > solve(D(f)(x)=0,x);Where the function has a denominator remember to check where the derivative is undefine as well as zero.

> g:=x->x^4/(7*x+5); > plot(g(x),x=-3..3,y=-3..3); > solve(D(g)(x)=0,x); > solve(D(g)(x)=infinity);The above command doesn't work so enter the denominator (using the

> solve(denom((D(g))(x)) = 0, x); > solve(1/D(g)(x)=0,x);So, the function has two critical values and the function has three critical values.

> D(g)(-5); D(g)(-1); D(g)(-1/2); D(g)(5);It is only the positive or negative sign that is important. The function increases on the interval , then decreases on the interval , and finally increases on the interval . Therefore, there is a maximum at and a minimum at .

> D[1,1](g)(-10/7); D[1,1](g)(-5/7); D[1,1](g)(0);(Note the

> solve(D(f)(x)=0,x);and the two endpoints and .Simply find the four y values and pick the highest and lowest value.

> f(-2.5); f(-1); f(1); f(1.5);Therefore, the ABSOLUTE extrema are (-2.5, -7.13) and (-1, 3).

- For the function
- A)
- Find ALL critical values and state them in text.
- B)
- Use the first derivative test to find intervals on which is increasing and intervals on which it is decreasing without looking at a plot of the function.

- Without plotting the function
, find all critical points and then classify each point as a relative maximum or a relative minimum using the second derivative test. Be sure to find corresponding values for each critical value. (You do not need to consider where the derivative is undefined here because there could never be a relative extrema at a point where is undefined.)
- Find the absolute extrema for the function
on the closed interval without plotting the function. Plot the
**derivative**of the function and use`fsolve`to help you find critical values.

2011-08-21