AIEEE 2010 (Offline)Question 1
Let $f:R\to R$ be defined by $f(x)=\begin{cases}k-2x,&\text{if }x\le-1\\2x+3,&\text{if }x>-1\end{cases}$
If $f$ has a local minimum at $x=-1$, then a possible value of $k$ is
Differentiation and its Application
Solution
Concept. Compare $f(-1)$ with the values just to the left and just to the right. Formulas used. (1) Local minimum at $c$: $f(c)\le f(x)$ for all $x$ near $c$Step 1 - Value. $f(-1)=k+2$. Step 2 - Left side. For $x<-1$: $k-2x>k+2$. Always fine. Step 3 - Right side. For $x>-1$: $2x+3>1$, and it takes values as close to $1$ as we like. By (1) we need $k+2\le1$, i.e. $k\le-1$. Step 4 - Option. Of the options only $k=-1$ satisfies this. Answer: (C)