Question

Let $f:R\to R$ be defined by $f(x)=\{\begin{array}{ll}k-2x,& \text{if }x\le -1\\ 2x+3,& \text{if }x>-1\end{array}$. If $f$ has a local minimum at $x=-1$ , then a possible value of $k$ is

• A. $-\frac{1}{2}$
• B. - 1
• C. 1
• D. 0

Answer 1

Answer: (B)

$f$ will be continuous at $x=-1$ if $\underset{x\to -1}{\lim }f(x)$
$=\underset{x\to -1+}{\lim }f(x)=f(-1)$
$\Rightarrow k+2=2(-1)+3=k+2$
$\Rightarrow k=-1$
For this value of $k$, $f$ is continuous at $x=-1$,
$f^{\prime }(-1)$ does not exist and $f^{\prime }(x)<0$ for $x<-1$
and $f^{\prime }(x)<0$ for $x>-1$
$\therefore f$ has a local minimum at $x=-1$.