Question

imit as x approaches 2 (f(x)-f(2))/(x-2) when f(x)=4x^2-2x+4

Answer 1

\(f(2)=4\times 2^2-2\times 2+4=16\)

Answer 2

so you are after
\[\lim_{x\to 2}\frac{4x^2-2x+4-16}{x-2}=\frac{4x^2-2x-12}{x-2}\]

Answer 3

if you replace \(x\) by \(0\) you get \(\frac{0}{0}\) so you know you can factor and cancel
factor as \[\frac{(x-2)(\text{something})}{x-2}=\text{something}\] then replace \(x\) by 2

Answer 4

rather i meant "if you replace \(x\) by 2"

Answer 5

can be its means that f'(2) if f(x) = 4x^2-2x+4
so, f'(x) = 8x-2
put x=2, to find f'(2)

Answer 6

so the limit should = 14

Answer 7

yes..