Find all relative extrema and points of inflection for the function $g(x)={\frac {2}{3}}x^{3}+x^{2}12x$. Be sure to give coordinate pairs for each point. You do not need to draw the graph.
Foundations:

Since our function is a polynomial, the relative extrema occur when the first derivative is zero. We then have two choices for finding if it is a local maximum or minimum:

Second Derivative Test: If the first derivative at a point $x_{0}$ is $0$, and the second derivative is negative (indicating it is concavedown, like an upsidedown parabola), then the point $\left(x_{0},f(x_{0})\right)$ is a local maximum.

On the other hand, if the second derivative is positive, the point $\left(x_{0},f(x_{0})\right)$ is a local minimum. You can also use the first derivative test, but it is usually a bit more work! For inflection points, we need to find when the second derivative is zero, as well as check that the second derivative "splits" on both sides.

Solution:
Step 1:

Find the first and second derivatives: Based on our function, we have

 $g\,'(x)\,=\,{\frac {2}{3}}\cdot 3x^{2}+2x12\,=\,2x^{2}+2x12.$

Similarly, from the first derivative we find

 $g\,''(x)\,=\,4x+2.$

Step 2:

Find the roots of the derivatives: We can rewrite the first derivative as

 $g\,'(x)\,=\,2x^{2}+2x12\,=\,2(x^{2}+x6)\,=\,2(x+3)(x2),$

from which it should be clear we have roots $2$ and $3$.

On the other hand, for the second derivative, we have

 $g\,''(x)\,=\,4x+2\,=\,4\left(x+{\frac {1}{2}}\right).$

This has a single root: $x={\frac {1}{2}}$.

Step 3:

Test the potential extrema: We know that $x=2,3$ are the candidates. We check the second derivative, finding

 $g\,''(2)\,=\,4\cdot 2+2\,>\,0,$

while

 $g\,''(3)\,=\,2(3)+2\,<\,0.$

Note that

 $g(2)\,=\,{\frac {2}{3}}(8)+424\,=\,{\frac {44}{3}},$

while

 $g(3)\,=\,{\frac {2}{3}}(27)+912(3)\,=\,27.$

By the second derivative test, the point $(2,g(2))=\left(2,{\frac {44}{3}}\right)$ is a relative minimum, while the point $(3,g(3))=(3,27)$ is a relative maximum.

Step 4:

Test the potential inflection point: We know that $g\,''(1/2)=0$. On the other hand, it should be clear that if $x<1/2$, then $g\,''(x)<0$. Similarly, if $x>1/2$, then $g\,''(x)>0$. Thus, the second derivative "splits" around $x=1/2$ (i.e., changes sign), so the point $\left(1/2,g(1/2)\right)$ is an inflection point.

Since

 $g\left({\frac {1}{2}}\right)\,=\,{\frac {2}{3}}\left({\frac {1}{8}}\right)+{\frac {1}{4}}12\left({\frac {1}{2}}\right)\,=\,{\frac {37}{6}},$

our inflection point is

 $\left({\frac {1}{2}},{\frac {37}{6}}\right).$

Final Answer:

There is a local minimum at $\left(2,{\frac {44}{3}}\right)$, a local maximum at $(3,27)$ and an inflection point at $\left({\frac {1}{2}},{\frac {37}{6}}\right).$

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