B brumby_3 Jul 2008 212 2 May 22, 2010 #1 If f(x) = sqrt(x+4), use a calculator to compute [f(4+h)-f(2)]/h for h = ±0.1, ±0.05 So do I just put f(x) in place of the f's when putting into my calculator and do the negative and positive for each ± correct?

If f(x) = sqrt(x+4), use a calculator to compute [f(4+h)-f(2)]/h for h = ±0.1, ±0.05 So do I just put f(x) in place of the f's when putting into my calculator and do the negative and positive for each ± correct?

D dwsmith MHF Hall of Honor Mar 2010 3,093 582 Florida May 22, 2010 #2 brumby_3 said: If f(x) = sqrt(x+4), use a calculator to compute [f(4+h)-f(2)]/h for h = ±0.1, ±0.05 So do I just put f(x) in place of the f's when putting into my calculator and do the negative and positive for each ± correct? Click to expand... \(\displaystyle \frac{f(4+h)-f(2)}{h}=\frac{\sqrt{(4+h)+4}-\sqrt{2+4}}{h}\) Reactions: brumby_3

brumby_3 said: If f(x) = sqrt(x+4), use a calculator to compute [f(4+h)-f(2)]/h for h = ±0.1, ±0.05 So do I just put f(x) in place of the f's when putting into my calculator and do the negative and positive for each ± correct? Click to expand... \(\displaystyle \frac{f(4+h)-f(2)}{h}=\frac{\sqrt{(4+h)+4}-\sqrt{2+4}}{h}\)

B brumby_3 Jul 2008 212 2 May 26, 2010 #3 So I need to plug the negative and positive of each number into the function to get the answer?