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Given $k(x) = \{2x^2 - 4$ for $x < 2$ and $-3x + 10$ for $x$ is equal to or less than $2$ . Is $k(x)$ continuous at $x=2$ ? Explain. If $j(x)=k(x+1)$ , what is the function for $j(x)$ ?

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Write and solve direct variation equations

Full solution

Q. Given

k(x) = \{2x^2 - 4

for

x < 2

and

-3x + 10

for

x

is equal to or less than

2

. Is

k(x)

continuous at

x=2

? Explain. If

j(x)=k(x+1)

, what is the function for

j(x)

?

Check Continuity at x= $2$ : Step $1$ : Analyze the function $k(x)$ at $x=2$ to check for continuity. $k(x)$ is defined as $\{2x^2 - 4 \text{ for } x < 2 \text{ and } -3x + 10 \text{ for } x \leq 2.$ To check continuity at $x=2$ , we need the limit from the left ( $x$ approaches $2$ from the left) to equal the function value at $x=2$ .Calculate the left-hand limit: $\lim_{x\to2^-} 2x^2 - 4$ .Substitute $x = 2$ : $x=2$ $0$ .Function value at $x=2$ (right-hand side of the definition): $x=2$ $2$ .Since both the left-hand limit and the function value at $x=2$ are equal, $k(x)$ is continuous at $x=2$ .
Define Transformation $j(x)$ : Step $2$ : Define $j(x)$ based on the transformation $j(x) = k(x+1)$ . Substitute $x+1$ for $x$ in each part of $k(x)$ . For $x+1 < 2$ (i.e., $x < 1$ ): $j(x) = 2(x+1)^2 - 4 = 2(x^2 + 2x + 1) - 4 = 2x^2 + 4x - 2$ . For $x+1 \leq 2$ (i.e., $j(x)$ $0$ ): $j(x)$ $1$ . Thus, $j(x)$ $2$ .

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