Question

se $y'= \lim_{x \to \00} \frac{f(x+delta . x) - f(x)}{delta . x}$, mostrar que
se $y=-3 x^{2} +5x+7$ então $y'=-6x+5$.

Answer 1

$\boxed{\boxed{f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} }}$

temos:
$f(x)=-3x^2+5x+7$

f(x+h) -> substitui o x em f(x) por (x+h)

então
$\boxed{\boxed{f(x+h)= -3(x+h)^2+5(x+h)+7}}$

lembrando que (a+b)² = (a²+2ab+b²)

resolvendo:

$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\\\f'(x)=\lim_{h \to 0} \frac{-3(x+h)^2+5(x+h)+7 - [-3x^2+5x+7]}{h} \\\\ f'(x)=\lim_{h \to 0} \frac{-3(x+h)^2+5(x+h)+\not7 +3x^2-5x-\not7]}{h} \\\\ f'(x)=\lim_{h \to 0} \frac{-3(x+h)^2+5(x+h)+3x^2-5x]}{h} \\\\ f'(x)=\lim_{h \to 0} \frac{-3(x+h)^2+\not5x+5h+3x^2-\not 5x]}{h}\\\\ f'(x)=\lim_{h \to 0} \frac{-3(x+h)^2+5h+3x^2}{h}\\\\ f'(x)=\lim_{h \to 0} \frac{-3(x^2+2xh+h^2)+5h+3x^2}{h}\\\\ f'(x)=\lim_{h \to 0} \frac{-\not3x^2-6xh-3h^2+5h+\not3x^2}{h}$

$f'(x)=\lim_{h \to 0} \frac{-6xh-3h^2+5h}{h} \\\\ f'(x)=\lim_{h \to 0} \frac{\not h(-6x-3h+5)}{\not h} \\\\ f'(x)=\lim_{h \to 0}( -6x+3h+5) = -6x+3*0 +5 = -6x+5$