Question

Question 5 and 6 please help me

Answer 1

Problem 5

The function is continuous for the given domain
`x \ge 6`

This is because y = (-5/6)x+5 is itself continuous, and any interval subset of this function is also continuous. We can plug in any real number that is equal to 6 or larger, and get some y output. If we plugged in x = 6, then we'd get

y = (-5/6)x+5

y = (-5/6)*6 + 5

y = -5+5

y = 0

This is the largest y value possible. Why? Because y = (-5/6)x+5 has a negative slope, so the graph is going downhill as you read it from left to right. As x gets bigger, y gets smaller. The smallest x value allowed in the domain produces the largest y value in the range. There is no smallest y value as the y values keep going down forever.

The range is therefore
`y \le 0`

In interval notation, you can write the range as
`(-\infty, 0]`. The square bracket indicates "include this endpoint as part of the range".

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Problem 6

The function is discrete for this given domain. The domain itself is a discrete list of values. We cannot plug in values between say 0 and 2. We can only substitute one of those values from the list given. Consequently, the y values will also be a list, and not an interval like problem 5 had.

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If you plugged in x = -4, then you should get...

y = (-1/2)*(-4)+2

y = 2+2

y = 4

So the input x = -4 lead the output y = 4

Repeat for x = -2

y = (-1/2)x+2

y = (-1/2)*(-2)+2

y = 1+2

y = 3

and the same for x = 0 as well

y = (-1/2)x+2

y = (-1/2)*0 + 2

y = 0 + 2

y = 2

and x = 2 also

y = (-1/2)x+2

y = (-1/2)*2 + 2

y = -1+2

y = 1

Finally, plug in x = 4

y = (-1/2)x+2

y = (-1/2)*4+2

y = -2+2

y = 0

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If we plugged each of these x values {-4, -2, 0, 2, 4} one at a time into the equation y = (-1/2)x+2, then we get this list of values {4, 3, 2, 1, 0}

Sorting the values from smallest to largest, we have this range {0, 1, 2, 3, 4}