+0

# Help asap???

+3
257
3

h(x)=x^2+4 (h∘h)(a)

Guest Sep 25, 2014

#3
+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
Sort:

#1
+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
+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
+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

### 31 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details