How do you write #f(x) = 3 - |2x + 3|# as a piecewise function?
Use the definition of the absolute value function:
and then simplify the domain restrictions.
Simplify the domain restrictions:
Use the distributive property to eliminate the ()s:
Equation [3] is the desired piece-wise function.
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(f(x) = \begin{cases} 3 - (2x + 3) & \text{if } 2x + 3 \geq 0 \ 3 -(-(2x + 3)) & \text{if } 2x + 3 < 0 \end{cases})
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[ f(x) = \begin{cases} 3 - (2x + 3) & \text{if } 2x + 3 \leq 0 \ 3 - (-(2x + 3)) & \text{if } 2x + 3 > 0 \end{cases} ]
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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