Question

Let $f\left(x\right)={(1+x)}^n-(1+nx)$, $x\in [-1,\infty )$. Then $f$

• A. has an absolute maximum at $x=0$
• B. has neither absolute maximum nor absolute minimum at $x=0$
• C. has an absolute minimum at $x=0$
• D. does not have absolute minimum at $x=0$

Answer 1

Answer: (C)

has an absolute minimum at $x=0$

$f\left(x\right)={(1+x)}^n-(1+nx)$
$\Rightarrow f^{\prime }\left(x\right)=n[{(1+x)}^{n-1}-1]$.
Now for $x=0$,$f^{\prime }\left(0\right)=0$
and for $x>0$,$f^{\prime }\left(x\right)>0$.
Thus $f$ inereases on $[0,\infty )$ i.e.$f\left(x\right)\ge f\left(0\right)=0$
For $-1\le x<0$, $0\le 1+x<1$ so ${(1+x)}^n-1<0$.
So $f$ decreases on $[-1,0)$ i.e. $f\left(x\right)\ge f\left(0\right)=0$ for $x\in [-1,0)$.
Hence $f$ has an absolute minimum at $x=0$.