Let f-R- R be defined by fx- k-2x- if x- -1 2x-3- if x-1- If f has a local minimum at x-1- then a possible value of k is-

The correct option is D 1 f has a local minimum at x= 1 f( 1)=k+2 When x gt; 1,f(x) gt;1 One of the possibility is k+2=1 k= ...

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Chain Rule of Differentiation

Let f:R→ R ...

Question

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

1

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Similar questions

f : R \to R

f (x) = {\begin{matrix} k - 2 x, i f x \leq - 1 2 x + 3, i f x > - 1 \end{matrix}

f

x = - 1

k

f : R \to R

f (x) = {\begin{matrix} k - 2 x, i f x \leq - 1 2 x + 3, i f x > - 1 \end{matrix}

k

f (x) = {\begin{matrix} k - 2 x, i f x \leq - 1 2 x + 3, i f x > - 1 \end{matrix}},

f (x) = {\begin{matrix} k - 2 x, i f x \leq 1 2 x + 3, i f x > - 1 \end{matrix}}

f

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{{cos}^{2} x - {sin}^{2} x - 1}{\sqrt{x + 1} - 1} & , if x \neq 0 k & , if x = 0 \end{matrix},

x \neq 0,

x = 0

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