How do you find f'(-5) using the limit definition given $f(x) = −2/(x + 1)$ ?

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Answer 1 · 2016-12-21T18:08:57

$f'(-5) = 1/8$

The limit definition is given by the formula $f'(x) = lim_(h-> 0)(f(x + h) - f(x))/h$ .

$f'(x) = lim_(h->0) (-2/(x + h + 1) - (-2/x + 1))/h$

$f'(x) = lim_(h-> 0) (-2/(x + h + 1) + 2/(x + 1))/h$

$f'(x) = lim_(h->0) ((-2(x + 1) + 2(x + h + 1))/((x + h + 1)(x + 1)))/h$

$f'(x) = lim_(h-> 0) ((-2x - 2 + 2x + 2h + 2)/((x + h + 1)(x + 1)))/h$

$f'(x) = lim_(h->0) (2h)/((x + h + 1)(x + 1)h)$

$f'(x) = lim_(h->0) (2)/((x + h + 1)(x + 1)$

We can evaluate now:

$f'(x) = 2/((x + 0 + 1)(x + 1))$

$f'(x) = 2/(x + 1)^2$

We now evaluate the derivative when $x= -5$ :

$f'(-5) = 2/(-5 + 1)^2$

$f'(-5) = 2/16$

$f'(-5) = 1/8$

Hopefully this helps!

How do you find f'(-5) using the limit definition given f(x) = −2/(x + 1)f(x) = −2/(x + 1)?