# How do you find the value of k where the 2 function continuous g(x)= |x^3| for -oo<x<k and x^2 for k<x<oo?

Oct 13, 2016

#### Answer:

$k \in \left\{- 1 , 0 , 1\right\}$

#### Explanation:

For $g \left(x\right) = \left\{\begin{matrix}| {x}^{3} | \mathmr{if} x \le k \\ {x}^{2} \mathmr{if} x \ge k\end{matrix}\right.$ to be continuous, we must have the piecewise components agree at the value at which they meet, i.e. we must have $| {k}^{3} | = {k}^{2}$.

We'll consider two cases:

Case 1: $k \ge 0$

$\implies | {k}^{3} | = {k}^{3}$

$\implies {k}^{3} - {k}^{2} = 0$

$\implies {k}^{2} \left(k - 1\right) = 0$

$\implies k = 0 \mathmr{and} k = 1$

Case 2: $k < 0$

$\implies | {k}^{3} | = - {k}^{3}$

$\implies - {k}^{3} - {k}^{2} = 0$

$\implies - {k}^{2} \left(k + 1\right) = 0$

$\implies k = - 1$

Together, we get three possible values for $k$ which will result in $g$ being continuous: $k \in \left\{- 1 , 0 , 1\right\}$