Says here, a function f is defined for all real numbers and has the following property: f(a + b) - f(a) = 3(a^2)b + 2b^2. f ' (x) is equal to what?
Basically, are a and b constants or something? How do I find f ' (x)?
Hello, LeoBloom!
Recall the defintion: .$\displaystyle f'(x) \;=\;\lim_{h\to0}\frac{f(x+h) - f(x)}{h} $A function $\displaystyle f$ is defined for all real $\displaystyle x$ and: .$\displaystyle f(a + b) - f(a) \;= \;3a^2b + 2b^2$
Find $\displaystyle f'(x)$
They have given us about half of this defintion: .$\displaystyle f(a+b) - f(a) \:=\:3a^2b - 2b^2$
Divide by $\displaystyle b\!:\;\;\frac{f(a+b) - f(a)}{b} \;=\;\frac{3a^2b + 2b^2}{b} \;=\;3a^2 + 2b$
Take the limit as $\displaystyle b\to0$
. . $\displaystyle \underbrace{\lim_{b\to0}\frac{f(a+b) - f(a)}{b}}_{\text{This is }f'(a)} \;=\;\lim_{b\to0}(3a^2 + 2b) \;=\;3a^2$
So we have: .$\displaystyle f'(a) \:=\:3a^2$
Therefore: .$\displaystyle \boxed{f'(x) \:=\:3x^2}$