# Find the value of k that makes the function f(x) = \left\{\begin{matrix} kx-1 & x \lt 2\\ ...

## Question:

Find the value of {eq}k {/eq} that makes the function {eq}f(x) = \left\{\begin{matrix} kx-1 & x \lt 2\\ (3-x)^2+1 & x \geq 2 \end{matrix}\right. {/eq} continuous.

## Continuity of a Function :

If {eq}\displaystyle \lim_{x \to c}f\left ( x \right )=f\left ( c \right ) {/eq}, then we say that {eq}f {/eq} is continuous at {eq}x=c {/eq}. That is, {eq}f {/eq} is continuous at {eq}c {/eq} if {eq}\displaystyle \lim_{x \to c^{-}}f\left ( x \right )=\lim_{x \to c^{+}}f\left ( x \right )=f\left ( c \right ) {/eq}.

## Answer and Explanation:

{eq}\begin{align*} f\left ( 2 \right )&=\left ( 3-2 \right )^2+1 \\ &=2 \\ \lim_{x \to 2^{-}}f\left ( x \right )&=\lim_{x \to 2^{-}}\left ( kx-1 \right ) \\ &=2k-1 \end{align*} {/eq}

Since {eq}f {/eq} is continuous, we have

{eq}\begin{align*} \lim_{x \to 2^{-}}f\left ( x \right )&=f\left ( 2 \right ) \\ \Rightarrow 2k-1&=2 \\ \Rightarrow 2k&=3 \\ k&=\frac{3}{2} \end{align*} {/eq}

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