How do you determine the intervals for which the function is increasing or decreasing given $f (x) = \frac{1}{x + 1} - 4$ $f(x)=1/(x+1)-4$ ?

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Answer 1 · 2017-09-08T13:03:29

The intervals of decreasing are $(- \infty, - 1) \cup (- 1, + \infty)$ $(-oo,-1) uu (-1,+oo)$

We need

$(1/x)'=-1/x^2$

$f(x)=1/(1+x)-4$

The domain of $f(x)$ is $D_f(x)=RR-{-1}$

Therefore, the derivative of $f(x)$ is

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

$AA x in D_f(x), f'(x)<0$

There are no critical values, and the variation table is

$color(white)(aaaa)$ $Interval$ $color(white)(aaaa)$ $(-oo,-1)$ $color(white)(aaaa)$ $(-1,+oo)$

$color(white)(aaaa)$ $sign f'(x)$ $color(white)(aaaaaaaa)$ $-$ $color(white)(aaaaaaaaaaa)$ $-$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaaaaaaaa)$ $↘$ $color(white)(aaaaaaaaaaa)$ $↘$

You can also, calculate the second derivative and determine the concavity.

graph{1/(1+x)-4 [-10, 10, -5, 5]}

How do you determine the intervals for which the function is increasing or decreasing given f(x)=1/(x+1)-4f(x)=1x+1−4f(x)=1/(x+1)-4?