Question

Let $f : R \,\to\, R$ be defined by $ f(x) = \begin{cases} k-2x , & \quad \text{if } x\,\leq\,-1\\ 2x+3, & \quad \text{if }x > \,-1\\ \end{cases} $. 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 $\displaystyle \lim_{x \to -1}f(x)$
$=\displaystyle \lim_{x \to -1+} f(x)=f(-1)$
$\Rightarrow k+2=2(-1)+3=k+2$
$\Rightarrow k=-1$
For this value of $k$, $f$ is continuous at $x = - 1$,
$f'(-1)$ does not exist and $f'(x) < 0$ for $x < - 1$
and $f'(x) < 0$ for $x > - 1$
$\therefore f$ has a local minimum at $x = - 1$.