Give a rule for the piecewise-defined function. Then give the domain and range.
- $Give a rule for the piecewise-defined function. Then give the domain and range. - A) f ( x ) = \left\{ \begin{array} { l l } x - 3 & \text { if } x \neq 3 \\ - 2 & \text { if } x = 3 \end{array} ; \right. Domain: ( - \infty , \infty ) , Range: ( - \infty , 3 ] \cup [ 3 , \infty ) B) f ( x ) = \left\{ \begin{array} { l l } 2 x - 3 & \text { if } x \neq 3 \\ - 2 & \text { if } x = 3 \end{array} ; \right. Domain: ( - \infty , \infty ) , Range: ( - \infty , 3 ) \cup ( 3 , \infty ) C) f ( x ) = \left\{ \begin{array} { l l } 2 x - 3 & \text { if } x \neq 3 \\ - 3 & \text { if } x = 3 \end{array} ; \right. Domain: ( - \infty , 3 ] \cup [ 3 , \infty ) , Range: ( - \infty , \infty ) D) f ( x ) = \left\{ \begin{array} { l l } 2 x - 3 & \text { if } x < 3 \\ 2 x + 3 & \text { if } x \geq 3 \end{array} ; \right. Domain: ( - \infty , 3 ) \cup ( 3 , \infty ) , Range: ( - \infty , \infty )$
A)

f ( x ) = \left\{ \begin{array} { l l } x - 3 & \text { if } x \neq 3 \\ - 2 & \text { if } x = 3 \end{array} ; \right.

Domain:

( - \infty , \infty )

, Range:

( - \infty , 3 ] \cup [ 3 , \infty )

f ( x ) = \left\{ \begin{array} { l l } 2 x - 3 & \text { if } x \neq 3 \\ - 2 & \text { if } x = 3 \end{array} ; \right.

Domain:

( - \infty , \infty )

, Range:

( - \infty , 3 ) \cup ( 3 , \infty )

f ( x ) = \left\{ \begin{array} { l l } 2 x - 3 & \text { if } x \neq 3 \\ - 3 & \text { if } x = 3 \end{array} ; \right.

Domain:

( - \infty , 3 ] \cup [ 3 , \infty )

, Range:

( - \infty , \infty )

f ( x ) = \left\{ \begin{array} { l l } 2 x - 3 & \text { if } x < 3 \\ 2 x + 3 & \text { if } x \geq 3 \end{array} ; \right.

Domain:

( - \infty , 3 ) \cup ( 3 , \infty )

, Range:

( - \infty , \infty )

Compute and Simplify the Difference Quotient f(x+h)−f(x)h,h≠0\frac { f ( x + h ) - f ( x ) } { h } , h \neq 0hf(x+h)−f(x)​,h=0

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Compute and Simplify the Difference Quotient $\frac { f ( x + h ) - f ( x ) } { h } , h \neq 0$

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