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h(x)=x^2+4 (h∘h)(a)

Guest Sep 25, 2014

Best Answer 

 #3
avatar+85675 
+8

We have ..... h(x) = x2 + 4

And I'm not sure, but I think this person might be wanting to evalute the composite function (h º h)(a)

This says to put "a" into f(x), and then put that result back into f(x) and evaluate again .................     (very odd, indeed !!!)

So we have......

h(a) = a2 + 4   and "plugging" this result back into h(x), we have

h(a2 + 4) = (a2 + 4)2 + 4  =   a4 + 8a2 + 20

And that's (h º h)(a)      ......  (If that was the original intent !! )

 

CPhill  Sep 26, 2014
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3+0 Answers

 #1
avatar
+3

Ok wow thats tough  I dont think thats possible. So no, this is not a helpful answer to your desperate plea. Sorry...

Guest Sep 25, 2014
 #2
avatar+92206 
+8

I am really guessing here - would someone like to verify please.

Plus I do not know how to use the notation properly.

 

h(x)=x^2+4 (h∘h)(a)

 

$$\\(x^2+4)^2+4 \\
$sub in a$\\
=(a^2+4)^2+4\\
=a^4+8a^2+16+4\\
=a^4+8a^2+20\\$$

Melody  Sep 26, 2014
 #3
avatar+85675 
+8
Best Answer

We have ..... h(x) = x2 + 4

And I'm not sure, but I think this person might be wanting to evalute the composite function (h º h)(a)

This says to put "a" into f(x), and then put that result back into f(x) and evaluate again .................     (very odd, indeed !!!)

So we have......

h(a) = a2 + 4   and "plugging" this result back into h(x), we have

h(a2 + 4) = (a2 + 4)2 + 4  =   a4 + 8a2 + 20

And that's (h º h)(a)      ......  (If that was the original intent !! )

 

CPhill  Sep 26, 2014

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