Spivak's Calculus: Using derivatives prove that if $n \geq 1$, then $(1+x)^n > 1+nx$ for $-10$. The following is a problem from Spivak's Calculus, Ch. 11



*Use derivatives to prove that if $n \geq 1$, then

$$(1+x)^n > 1+nx, \text{ for } -1<x<0 \text{ and } x>0$$
(notice that equality holds for $x=0$)

The solution in the solution manual is a bit terse

Let $g(x)=(1+x)^n-(1+nx)$. Then $g(0)=0$, but $$g'(x)=n(1+x)^{n-1}-n\tag{1}$$
Since $n-1 \neq 0$ this means that
$$\begin{align}g'(x) & < 0 \text{ for } -1<x<0, \\ & >0 \text{ for }
 x>0 \end{align}\tag{2}$$
Thus $g(x)>0$ for $-1<x<0$ and $x>0$

I'd like to fill in the the steps in more detail. Also, my solution differs from the solution above and I'd like to know why.
Consider, for example, the case $n=2$. Then
$$g(x)=(1+x)^2-(1+2x)$$
$$g'(x)=2(1+x)-2$$
$$g(-1.5)=(1-1.5)^2-(1+2\cdot (-1.5))=\frac{1}{4}+2>0$$
$$g'(-1.5)=2(1-1.5)-2=-3$$
Why doesn't the solution account for the interval $(-2,-1)$?
Here is what I mean.
Consider $(1)$
$$g'(x)=n[(1+x)^{n-1}-1]$$
$$g'(x)<0 \implies (1+x)^{n-1}-1<0$$
$$\implies (1+x)^{n-1}<1$$
If $n-1$ is odd, then
$$1+x<1 \implies x<0$$
If $n-1$ is even, then
$$1+x<1 \implies x<0$$ or $$1+x>-1 \implies x>-2$$
Therefore, for any $n\geq 1$

*

*if $n$ is even, then $g$ is decreasing on $(-\infty,0)$ and increasing on $(0,\infty)$. $0$ is a local minimum.

*if $n$ is odd, then $g$ is decreasing on $(-2,0)$ and increasing on $(\infty,-2)$ and $(0,\infty)$. Therefore, there is another root of $g(x)$ other than $0$. $0$ is a local minimum, $-2$ is a local maximum.

Therefore

*

*if $n$ is even, $g(x)>0$ for all $x \neq 0$

*if $n$ is odd, $g(x)>0$ for all $x$ above some value $x_0$ such that $g(x_0)=0$, excluding $0$.

Is my reasoning incorrect or is the solution manual incorrect?


Kind of proof without derivative ... 
$(1+x)^n > 1+ nx$ means that $(1+x)^n - 1^n > nx$
Now, factoring the right hand side we get
$x[1 + (1+x) + (1+x)^2 + ... +(1+x)^{n-1}) > nx$
so, when $x> 0$ this reduces to:
$[1 + (1+x) + (1+x)^2 + ... + (1+x)^{n-1}] > n$
which is of course true.
But if $x$ is negative (and $>-1$) dividing by $x$ changes sign so
we get
$[1 + (1+x) + (1+x)^2 + ... + (1+x)^{n-1}] < n$
which also holds.
