Investigating functions

You now know the individual tools: roots, derivatives, maxima and minima, inflection points. In this topic they are assembled into a procedure. The benefit is concrete: at the end you can sketch a function you have never seen before, without filling in a single table of thirty values.

The plan

The order is not arbitrary, it goes from coarse to fine. First the domain: where does the function exist at all? Second the behaviour at the edges: where does the curve run far to the left and far to the right? Third the intersections with the axes. Fourth the extreme points via $f'$. Fifth the inflection points via $f''$. And sixth you assemble everything into a sketch.

One look is worth taking in advance: symmetry. If only even exponents appear, the graph is symmetric about the $y$ axis, as for $x^4 - 3x^2$. If only odd ones appear, it is point-symmetric about the origin, as for $x^3 - 4x$. That halves your work, because whatever you find for positive $x$ also holds, mirrored, for negative ones.

One example from start to finish

Take $f(x) = x^3 - 3x$. The domain is all real numbers, because there is neither a fraction nor a root. Only odd exponents, so point symmetry about the origin. The degree is odd and the leading coefficient positive, so the curve comes up from the bottom left and runs to the top right.

Roots: $x^3 - 3x = x(x^2 - 3) = 0$ gives $x = 0$ and $x = \pm \sqrt{3} \approx \pm 1.73$. Extreme points: $f'(x) = 3x^2 - 3 = 0$ gives $x = \pm 1$. With $f''(x) = 6x$ we get $f''(1) = 6 > 0$, so a minimum at $(1|-2)$, and $f''(-1) = -6 < 0$, so a maximum at $(-1|2)$. Inflection point: $6x = 0$ gives $x = 0$, and there $f(0) = 0$.

That fixes the curve without your having to calculate a single extra point: it comes from below, rises to the maximum $(-1|2)$, falls through the origin to the minimum $(1|-2)$ and then rises forever. The inflection point at the origin is exactly where it falls most steeply.