Question

f(x)=ax^2+2 if x<=2, x+20a+8/x^2+1 if x>2 find a so that f'(2) exists

Answer 1

Left hand derivative and right hand derivative have to be equal..

Answer 2

?? did i misread the function
f(x) = x +20a +8/x^2

Answer 3

oh i didn't see the "+1"

Answer 4

sorry

Answer 5

is that one fraction ?
please clarify functions with parenthesis or use equation editor

Answer 6

ok clarified

Answer 7

ok to find that derivative use quotient rule
\[\rightarrow \frac{f'g - fg'}{g^{2}}\]
f = x+20a +8 .................. f' = 1
g = x^2 +1 ..................... g' = 2x
\[= \frac{(x^{2}+1)-2x(x+20a+8)}{(x^{2}+1)^{2}}\]
\[=\frac{1-x^{2}-40ax-16x}{(x^{2}+1)^{2}}\]
Now plug in 2 for x, and set equal to 2ax
\[\rightarrow 4a = \frac{1-4-80a-32}{25}\]
solve for a from there