Convert the following absolute value function into piece wise function.
$f (x) = | x^{2} - 4 x + 5 |$

{\begin{matrix} - x^{2} + 4 x + 5 & i f x \leq - 1 o r x \geq 5 (x^{2} - 4 x - 5) & i f - 1 < x < 5 \end{matrix}

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{\begin{matrix} x^{2} + 4 x + 5 & i f x \leq - 1 o r x \geq 5 - (x^{2} - 4 x - 5) & i f - 1 < x < 5 \end{matrix}

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{\begin{matrix} x^{2} - 4 x - 5 & i f x \leq - 1 o r x \geq 5 - (x^{2} - 4 x - 5) & i f - 1 < x < 5 \end{matrix}

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None of these

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Solution

The correct option is C ${\begin{matrix} x^{2} - 4 x - 5 & i f x \leq - 1 o r x \geq 5 - (x^{2} - 4 x - 5) & i f - 1 < x < 5 \end{matrix}$
Correction: Expression is $x^{2} - 4 x - 5$

$| x^{2} - 4 x - 5 | = | x (x - 5) + (x - 5) | = | (x + 1) (x - 5) |$

Clearly

$f (x) = x^{2} - 4 x - 5$ when $x \geq 5$ or $x \leq - 1$ \

$= - (x^{2} - 4 x - 5)$ when $- 1 < x < 5$

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Similar questions

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