If f(x)= x^2,&when x 0 Find: 1. f (1 2 ) 2. f(-2) 3. f(1) 4. f(3) 5. f(-3)

We have, f(x)= x^2,&when x 0 1. f (1 2 )=1 2 2. f(-2)=(-2)^2=4 3. f(1)=1 1 =1 4. f(3)=1 3 5. f(-3)= does not exist because 3 domain (f).

If f(x)= x^2,&when x<0 ,&when 0 <1 1 x ,&when x>0 Find: 1. f (1 2…

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Question

If $\text{f(x)}=\begin{cases}\text{x}^2,&\text{when }\text{ x}<0\\\text{x},&\text{when }\ 0\leq\text{x}<1\\\frac{1}{\text{x}},&\text{when }\text{ x}>0\end{cases}$
Find:

$\text{f}\Big(\frac{1}{2}\Big)$
$\text{f}(-2)$
$\text{f}(1)$
$\text{f}(\sqrt{3})$
$\text{f}(\sqrt{-3})$

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Answer

We have,
$\text{f(x)}=\begin{cases}\text{x}^2,&\text{when }\text{ x}<0\\\text{x},&\text{when }\ 0\leq\text{x}<1\\\frac{1}{\text{x}},&\text{when }\text{ x}>0\end{cases}$

$\text{f}\Big(\frac{1}{2}\Big)=\frac{1}{2}$
$\text{f}(-2)=(-2)^2=4$
$\text{f}(1)=\frac{1}{1}=1$
$\text{f}(\sqrt{3})=\frac{1}{\sqrt{3}}$
$\text{f}(\sqrt{-3})=$ does not exist because $\sqrt{3}\notin$ domain $(f).$

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