Let f(x) = sqrt(x^2 + 16) - 5
Find f(1+a).

Question

thesmartone · Answer

plug in 1+a in place of x

anonymous · Answer

yes i ended up with "a" which is wrong

anonymous · Answer

not sure what I did wrong

anonymous · Answer

$$f(x) = \sqrt{x^2 + 16} - 5$$
$$f(1 + a) = \sqrt{(1 + a)^2 + 16} - 5 = ~a? ~~really?$$
@PeterSheperd

alexandervonhumboldt2 · Answer

$$f(1+a)=\sqrt{(1+a)^2+16}-5$$

thesmartone · Answer

$$f(1+a)=\sqrt{(1+2a + a^2+16}-5$$

alexandervonhumboldt2 · Answer

DIRECT ANSWER

thesmartone · Answer

I haven't even completed it, lol

anonymous · Answer

ah i thought you could just cancel the exponent and take square root of 16

anonymous · Answer

Smart One  ~__~   

@TheSmartOne

alexandervonhumboldt2 · Answer

not true lol

alexandervonhumboldt2 · Answer

i would say that this cannot be simplified any further

alexandervonhumboldt2 · Answer

all wut you can do is simplifying 1+16 and that is the end

thesmartone · Answer

I just did one step, I didn't even do it all .-.

nnesha · Answer

$\color{blue}{\text{Originally Posted by}}$ @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