Question

Se f(x)= x²+2x, achar
f(a+h)-f(a),diferente de 0

Answer 1

$f(\cdot)=(\cdot)^{2}+2(\cdot)\begin{cases}f(a)=a^{2}+2a\\f(a+h)=(a+h)^{2}+2(a+h)\end{cases}$

Desenvolvendo $f(a+h)$:

$f(a+h)=(a+h)^{2}+2(a+h)\\\\f(a+h)=(a^{2}+2\cdot a\cdot h+h^{2})+(2a+2h)\\\\f(a+h)=a^{2}+2ah+h^{2}+2a+2h\\\\f(a+h)=a^{2}+2ah+2h+2a+h^{2}\\\\f(a+h)=a^{2}+2a+h(2a+2+h)$

Então:

$\dfrac{f(a+h)-f(a)}{h}=\dfrac{a^{2}+2a+h(2a+2+h)-(a^{2}+2a)}{h}\\\\\\\dfrac{f(a+h)-f(a)}{h}=\dfrac{a^{2}+2a-a^{2}-2a+h(2a+2+h)}{h}\\\\\\\dfrac{f(a+h)-f(a)}{h}=\dfrac{h(2a+2+h)}{h}$

Como $h\neq0$, podemos cortar h:

$\boxed{\boxed{\dfrac{f(a+h)-f(a)}{h}=2a+2+h}}$