x + 2 \enspace \enspace if \enspace x < 0\\

1 – x \enspace \enspace if \enspace x > 0

\end{matrix}\right.\), Concluate f(-1), f(3) and sketch graph.

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Let f(x) = \(\left\{\begin{matrix}

x + 2 \enspace \enspace if \enspace x < 0\\

1 – x \enspace \enspace if \enspace x > 0

\end{matrix}\right.\)

Then

f(-1) = (-1) + 2 = 1 *(We have used x + 2 since -1 < 0, and when x < 0 then function f(x) = x + 2)*

f(3) = 1 – 3 = -2 *(We** h**ave** used 1 – x since 3 > 0, and when x > 0 then function f(x) = 1 – x)*

For graph of f(x), given that

f(x) = x + 2 for x < 0

i.e y = x + 2

Which is the linear equation, so is a straight line and it takes

x |
-1 | -2 |

y |
1 | 0 |

This means y = x + 2 passes through (-1, 1) and (-2, 0).

Also, given that

f(x) = 1 – x for x > 0

i.e y = 1 – x

Which is the linear equation, so is a straight line and it takes

x |
1 | 2 |

y |
0 | -1 |

This means y = x + 2 passes through (1, 0) and (2, -1).

With these information, the graph of the equation is as:

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