Originally Posted by

**MarkFL** Consider that:

$\displaystyle (x+1)^3=x^3+3x^2+3x+1$

$\displaystyle -3(x+1)^2=-3x^2-6x-3$

Adding the two expressions we get:

$\displaystyle x^3-3x-2$

And so:

$\displaystyle g(x)=f(x+1)-3=x(x^2-3)$

We can then see that $\displaystyle g(x)$ must be:

Increasing on: $\displaystyle (-\infty,-a)$

Decreasing on: $\displaystyle (-a,a)$

Increasing on: $\displaystyle (a,\infty)$

Where $\displaystyle 0<a<\sqrt{3}$

We can further deduce that $\displaystyle g(x)$ has a local maximum at:

$\displaystyle (-a,g(-a))=(-a,-g(a))$

And a local minimum at:

$\displaystyle (a,g(a))$

Now, suppose in our attempt to find $\displaystyle a$, we try:

$\displaystyle a=1$

As this is the only integral value allowed. We find:

$\displaystyle g(1)=-2$

Now, we can look at (where $\displaystyle 0<h<\sqrt{3}-1$):

$\displaystyle g(1+h)=(1+h)^3-3(1+h)=1+3h+3h^2+h^3-3-3h=-2+3h^2+h^3>-2$

Likewise:

$\displaystyle g(1-h)=(1-h)^3-3(1-h)=1-3h+3h^2-h^3-3+3h=-2+3h^2-h^3>-2$

We really "lucked out" with our choice of $\displaystyle a$.

And so we can conclude that $\displaystyle g(x)$ has a local maximum at $\displaystyle (-1,2)$ and a local minimum at $\displaystyle (1,-2)$. Transforming these coordinates back to $\displaystyle f(x)$ we then have:

Local maximum at $\displaystyle (-1+1,2+3)=(0,5)$

Local minimum at $\displaystyle (1+1,-2+3)=(2,1)$

Increasing on $\displaystyle (-\infty,0)$

Decreasing on $\displaystyle (0,2)$

Increasing on $\displaystyle (2,\infty)$