Question

I have a table and I need to evaluate this (af+3g)(1). How could I solve this?

Answer 1

(2f + 3g)(1) = 2f(1) + 3g(1)
Look up f(1) and g(1) in the table and substitute.

Answer 2

so 60*(1)+(-7)*(1)=53?

Answer 3

What is f(1) from the table?

Answer 4

50

Answer 5

correct. What is g(1) from the table?

Answer 6

-5

Answer 7

Yes. So 2f(1) + 3g(1) = 2(50) + 3(-5) = 100 - 15 = 85

Answer 8

ohhh okay I got confused with the numbers in front of the letters. How do I do ( f of g)(-1)?

Answer 9

(f o g)(-1) = f(g(-1))
Look up g(-1) in the table and then put that value in f(g(-1)) and look up f(?)

Answer 10

First what is g(-1) from the table?

Answer 11

-3

Answer 12

correct. So f(g(-1)) = f(-3).
What is f(-3) from the table?

Answer 13

-1

Answer 14

That is g(-3). We need f(-3).

Answer 15

f(-3)=10

Answer 16

That is your answer.

Answer 17

what?

Answer 18

10 is your final answer.

Answer 19

Im confused how you got that

Answer 20

(f o g)(-1) = f(g(-1)) = f(-3) = 10.

Answer 21

From the table, g(-1) = -3
So f(g(-1)) = f(-3)
From the table f(-3) = 10

So the answer to the question is 10.

Answer 22

ohhh okay I see where you got that

Answer 23

I have another question if you dont mind helping me

Answer 24

go ahead.

Answer 25

Evaluate f(-2).

The function has two domains. It is 3x -2 if x <= 1 and 1/2x^3 if x > 1
We are interested in find f(x) when x = -2.
Which function should we use?

Answer 26

the first one?

Answer 27

Yes. Since -2 <= 1, we should use the first one.
So f(x) = 3x - 2
f(-2) = 3(-2) - 2 = -6 - 2 = -8

Answer 28

How do you graph this without a y value?

Answer 29

The f(x) is the y value.
Since there are two domains with different functions, you will have to graph them in two parts. First part, x <= 1, f(x) = y = 3x - 2.
Can you graph this straight line?

Answer 30

Im not really understanding this

Answer 31

Here there are two functions, each in its own domain.

For x <= 1, f(x) = 3x - 2. Let us graph this first. Remember: (1) f(x) is same as y and (2) the straight line should not go past x = 1 because the straight line is good only for x <= 1. So draw a line up to and including the point x = 1.

Answer 32

Yes. But don't go past x = 1. Stop at x = 1. The line can be extended downward as much as you want but upward it has to stop at (1,1).

Answer 33

ohh okay thank you

Answer 34

Are you good at word problems?

Answer 35

You are not done here. There is a second part to the graph where for x > 1 f(x) = 1/2 * x^3. You have to graph this too on the same graph.

Answer 36

awwh):

Answer 37

You can put a few values for x:
x x^3 / 2
1 1/2
2 8/2 = 4
3 27/2 = 13.5

But the point x = 1 belongs to the previous straight line part of the graph and not on the curve x^3/2. So at x = 1 put an open circle on this curve indicating that x = 1 is not included in this graph. In the previous graph, at x = 1, put a closed circle to indicate the point (1,1) belongs to the straight line and x = 1 is included in that straight line.