Local extrema/ relative maxima and minima

For the function y = x^3 - 3x^2 - 9x +15?

I solved and found an absolute max at (-1, 20) and absolute min at (3, -12). Does it even have a local max/min?

Since this is a cubic, which goes in both directions ( vertically) forever, it has no absolute max or min
Y'=3x^2-6x-9=0

3(x^2-2x-3)=0

3(x-3)(x+1)=0

X=3 or -1

(3,-12) is a relative min or local min

(-1,20) is a relative max or local max

A relative or local extreme indicates the turning points for the curve.

Good luck!

You've got it reversed. Local or relative extrema are just all the small humps. GLOBAL or ABSOLUTE extrema are the largest or smallest values that a function can take.

So the extrema happen at x=-1 and x=3, just like you found. Those are LOCAL extrema. They're only extreme in the LOCAL area around -1 and 3, respectively.

Cubics (and ALL odd degree polynomails) can never have absolute extrema if the domain is unrestricted. That's because the "end behavior" of y = x^3 shoots off to (inf, inf) and (-inf, -inf). I'm abusing notation here, but you can write "limit as x-->inf of x^3 = inf".

So, since y = x^3 can reach both -inf and +inf, it will have neither a min nor a max. Scaling it won't change it...2x^3 does the same thing. Flipping it upside down won't change it...you get (-inf, inf) and (inf, -inf) for the limits of y = -x^3. And adding smaller degrees won't change end behavior. The end behavior of x^3 - 3x^2 is controlled by the x^3. For sufficiently large x, the 3x^2 won't matter.

Even degree polynomials are a different story. They MUST have either a global max OR a global min (if the domain for x is unrestricted). For example, y = x^2 has an absolute/global min at x=0.

That's because x^2 reaches off to (-inf, +inf) and (+inf, +inf). Again, abusing some terms and notation, when you take a very large negative number like -99999999999 and square it (or raise it to ANY even power), it becomes very large and POSITIVE. Thus, x^2 has no absolute max, but does have an absolute min.