Maximum Value of $g(x)=(8+x)^3(8-x)^4$ I think there can not be a maximum value of this, as if I plug $x=$1000, it will increase the value of the function in leaps and bounds. The answer says that the maximum value will occur at $x=-8/7$. What am I missing here?

We have that
$$g(x)=(8+x)^3(8-x)^4\implies g'(x)=3(8+x)^2(8-x)^4-4(8+x)^3(8-x)^3=0$$
$$3(8+x)^2(8-x)^4=4(8+x)^3(8-x)^3$$
which is true when


*

*$x=\pm 8$

*$3(8-x)=4(8+x) \implies 7x=-8 \implies x=-\frac87$
and the latter is a local maximum.
