Question

Calcule a derivada de f(x)=x>2-4x+20 no ponto x0=5 usando limites

Answer 1

$f(x) = x^{2} - 4x + 20\\\\\\f'(5) =  \lim_{x \to 5} \frac{f(x)-f(5)}{x-5}\\\\f'(5) =  \lim_{x \to 5} \frac{x^{2}-4x+20-(5^{2}-4*5+20)}{x-5}\\\\f'(5) =  \lim_{x \to 5} \frac{x^{2}-4x+20-25}{x-5}\\ \\f'(5) =  \lim_{x \to 5} \frac{x^{2}-4x-5}{x-5}\\\\f'(5) =  \lim_{x \to 5} \frac{(x+1)(x-5)}{(x-5)}\\  \\f'(5) =  \lim_{x \to 5} x + 1\\\\f'(5) =  \lim_{x \to 5} 5 + 1\\ \\f'(5) = 6$

Espero ter ajudado :)

PS.: Desconsidera essas letras '$Â$[tex][/tex]'

Answer 2

f(x)=x”-4x+20
f’(x)=2x-4 pela regra da derivação
f’(5)=2.5-4 = 6