Question

So I'm given this:
f(X)={(3,5), (2,4), (1,7)}
g(X)=sqrt{x-3}
h(x)= {(3,2), (4,30, (1,6)}
k(x)= x^2+5

And then asked:
(f+h)(1)=?

what do I do?

Answer 1

find \(f(1)\) then find \(h(1)\) and then add

Answer 2

do you know what \(f(1)\) is? "no" is a fine answer, i am just asking
if you don't know i will show you what it is

Answer 3

i was thinking f(h(1))

Answer 4

i dont know

Answer 5

it is not asking for \(f(h(1))\) it is asking for \(f(1)+h(1)\)

Answer 6

\[f(x)=\{(3,5), (2,4), (1,7)\}\] to find \(f(1)\) find the ordered pair where the first coordinate is \(1\)
then \(f(1)\) is the second coordinate of that pair

Answer 7

nothing to compute here, we see \((1,7)\) and know that \(f(1)=7\)

Answer 8

\[h(x)= \{(3,2), (4,30, (1,6)\}\] what is \(h(1)\)?

Answer 9

6!

Answer 10

exactly!

Answer 11

so what you are being asked is really "what is \(7+6\) ? "

Answer 12

awesome, I understand. how about (f (a circle symbol) h) 3

Answer 13

\[f\circ h(x)\]?

Answer 14

yes! do i mulitply the 3?

Answer 15

oh i see \(f\circ h(3)\)

Answer 16

oh hell no

Answer 17

that is not a multiplication
that means
\[f(h(3))\]

Answer 18

work always from the inside out
the first thing you need to find it \(h(3)\)

Answer 19

\[h(x)= \{(3,2), (4,30, (1,6)\}\] what is \(h(3)\) ?

Answer 20

(3,2)

Answer 21

\(h(3)\) is a number, not an ordered pair
\((3,2)\) is an ordered pair

Answer 22

the first coordinate is \(3\) and the second coordinate is \(h(3)\)

Answer 23

recall that you said \(h(1)=6\) right? it was a number, not \((1,6)\)

Answer 24

so what is \(h(3)\) ?

Answer 25

2?

Answer 26

yes!

Answer 27

the ordered pair is \((3,2)\) so \(h(3)=2\) we are not done yet though

Answer 28

\[f(h(3))=f(2)\] since we know \(h(3)=2\) now you have to find \(f(2)\)

Answer 29

4?

Answer 30

\[f(x)=\{(3,5), (2,4), (1,7)\}\] yes

Answer 31

that is your answer
lets put it together all in one line

Answer 32

\[f\circ h(3)=f(h(3))=f(2)=4\]

Answer 33

this is the last questions ill ask but what if theres a negative one exponent?
f^-1(x)=?

Answer 34

\[f^{-1}(x)\] is the "inverse function" of \(f\)
if your function is given by ordered pairs, all you have to do is switch the coordinates of each ordered pair
i.e. put the second one first and the first one second

Answer 35

is it clear what i mean by that?

Answer 36

oh I see, and if i have K^-1(x)= is it y^2+5

Answer 37

let me check

Answer 38

ok for \(f^{-1}\) since
\[f(x)=\{(3,5), (2,4), (1,7)\}\] you have
\[f^{-1}(x)=\{(5,3), (4,2), (7,1)\}\] see what i did for that?

Answer 39

yes thats what i got, but does that apply to k^-1(X)? do i change X^2 to its inverse y^2? or its there more to it?

Answer 40

you can write
\[k(x)= x^2+5\]replace it by
\[y=x^2+5\] then switch \(x\) and \(y\) because that is what the inverse function does, and write
\[x=y^2+5\] then solve for \(y\) in two steps

Answer 41

\[x=y^2+5\]
\[x-5=y^2\]
\[\pm\sqrt{x-5}=y\]

Answer 42

as you can see the inverse of \(k\) is not a function, because of that \(\pm\) out front
the reason the inverse is not a function is because \(k(x)=x^2+5\) is not "one to one"
for example \(k(2)=9\) and \(k(-2)=9\)as well

Answer 43

so the inverse is not a function because \((2,9)\) and \((-2,9)\) are both on the graph of \(k\) so the graph of the inverse would have both \((9,2)\) and \((9,-2)\)
a function cannot have that

Answer 44

thank you so much for your help!

Answer 45

yw