Question

How do you sketch the general shape of `f(x)=x^5-3x^3+2x+4` using end behavior?

Answer 1

See graph and explanation. The second graph reveals turning points and points of inflexion (POI).

Explanation:

graph{x^5-3x^3+2x^2+4 [-39.86, 39.8, -20.43, 19.43]}

`f = x^5(1-3/x^2+2/x^4+4/x^5) to +-oo`, as `x to +-oo`.

`f'=3x^4-6x^2+2`, giving turning points?POI, at its

zeros x = +-1.24 and +-0.51, nearly,

`f''= 12x(x^2-1)=0`, at `x = 0 and x = +-1`.

`f'''=12(3x^2-1) ne 0`, at these points.

So, the POI are at `x= 0, +-1/sqrt3`.

The second graph locates the turning points and POE that could not

be located in the first graph

.graph{x^5-3x^3+2x^2+4[-1.5, 1.5, -20.43, 19.43]}