anonymous
  • anonymous
Let f(x) = sqrt(x^2 + 16) - 5 Find f(1+a).
Mathematics
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SOLVED
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chestercat
  • chestercat
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TheSmartOne
  • TheSmartOne
plug in 1+a in place of x
anonymous
  • anonymous
yes i ended up with "a" which is wrong
anonymous
  • anonymous
not sure what I did wrong

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anonymous
  • anonymous
\[f(x) = \sqrt{x^2 + 16} - 5\] \[f(1 + a) = \sqrt{(1 + a)^2 + 16} - 5 = ~a? ~~really?\] @PeterSheperd
AlexandervonHumboldt2
  • AlexandervonHumboldt2
\[f(1+a)=\sqrt{(1+a)^2+16}-5\]
TheSmartOne
  • TheSmartOne
\[f(1+a)=\sqrt{(1+2a + a^2+16}-5\]
AlexandervonHumboldt2
  • AlexandervonHumboldt2
DIRECT ANSWER
TheSmartOne
  • TheSmartOne
I haven't even completed it, lol
anonymous
  • anonymous
ah i thought you could just cancel the exponent and take square root of 16
anonymous
  • anonymous
Smart One ~__~ @TheSmartOne
AlexandervonHumboldt2
  • AlexandervonHumboldt2
not true lol
AlexandervonHumboldt2
  • AlexandervonHumboldt2
i would say that this cannot be simplified any further
AlexandervonHumboldt2
  • AlexandervonHumboldt2
all wut you can do is simplifying 1+16 and that is the end
TheSmartOne
  • TheSmartOne
I just did one step, I didn't even do it all .-.
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @PeterShepherd ah i thought you could just cancel the exponent and take square root of 16 \(\color{blue}{\text{End of Quote}}\) nope you can't do that \[\rm \sqrt{(1+a)^2+16}-5 \cancel{=} \sqrt{(1+a)^2}+\sqrt{16} \] there is a plus sign between x^2 and 16 so it's not possible to splt them into two parts it's not like a multiplication

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