anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
find \(f(1)\) then find \(h(1)\) and then add
anonymous
  • anonymous
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
anonymous
  • anonymous
i was thinking f(h(1))

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anonymous
  • anonymous
i dont know
anonymous
  • anonymous
it is not asking for \(f(h(1))\) it is asking for \(f(1)+h(1)\)
anonymous
  • anonymous
\[f(x)=\{(3,5), (2,4), (1,7)\}\] to find \(f(1)\) find the ordered pair where the first coordinate is \(1\) then \(f(1)\) is the second coordinate of that pair
anonymous
  • anonymous
nothing to compute here, we see \((1,7)\) and know that \(f(1)=7\)
anonymous
  • anonymous
\[h(x)= \{(3,2), (4,30, (1,6)\}\] what is \(h(1)\)?
anonymous
  • anonymous
6!
anonymous
  • anonymous
exactly!
anonymous
  • anonymous
so what you are being asked is really "what is \(7+6\) ? "
anonymous
  • anonymous
awesome, I understand. how about (f (a circle symbol) h) 3
anonymous
  • anonymous
\[f\circ h(x)\]?
anonymous
  • anonymous
yes! do i mulitply the 3?
anonymous
  • anonymous
oh i see \(f\circ h(3)\)
anonymous
  • anonymous
oh hell no
anonymous
  • anonymous
that is not a multiplication that means \[f(h(3))\]
anonymous
  • anonymous
work always from the inside out the first thing you need to find it \(h(3)\)
anonymous
  • anonymous
\[h(x)= \{(3,2), (4,30, (1,6)\}\] what is \(h(3)\) ?
anonymous
  • anonymous
(3,2)
anonymous
  • anonymous
\(h(3)\) is a number, not an ordered pair \((3,2)\) is an ordered pair
anonymous
  • anonymous
the first coordinate is \(3\) and the second coordinate is \(h(3)\)
anonymous
  • anonymous
recall that you said \(h(1)=6\) right? it was a number, not \((1,6)\)
anonymous
  • anonymous
so what is \(h(3)\) ?
anonymous
  • anonymous
2?
anonymous
  • anonymous
yes!
anonymous
  • anonymous
the ordered pair is \((3,2)\) so \(h(3)=2\) we are not done yet though
anonymous
  • anonymous
\[f(h(3))=f(2)\] since we know \(h(3)=2\) now you have to find \(f(2)\)
anonymous
  • anonymous
4?
anonymous
  • anonymous
\[f(x)=\{(3,5), (2,4), (1,7)\}\] yes
anonymous
  • anonymous
that is your answer lets put it together all in one line
anonymous
  • anonymous
\[f\circ h(3)=f(h(3))=f(2)=4\]
anonymous
  • anonymous
this is the last questions ill ask but what if theres a negative one exponent? f^-1(x)=?
anonymous
  • anonymous
\[f^{-1}(x)\] 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
anonymous
  • anonymous
is it clear what i mean by that?
anonymous
  • anonymous
oh I see, and if i have K^-1(x)= is it y^2+5
anonymous
  • anonymous
let me check
anonymous
  • anonymous
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?
anonymous
  • anonymous
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
\[x=y^2+5\] \[x-5=y^2\] \[\pm\sqrt{x-5}=y\]
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
thank you so much for your help!
anonymous
  • anonymous
yw

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