Question

How do you determine the intervals for which the function is increasing or decreasing given `f(x)=(x^2+5)/(x-2)`?

Answer 1

See the explanation.

Explanation:

By actual division,

f(x) = y = x+2+9/(x-2)

`y'=1-9/(x-2)^2=0`, when `(x-2)^2=9 to x = -1 and 5.`

For `x in (-oo, -1], y uarr,` from `-oo to -2`..

For `x in [-1, 2), y darr to -oo`.

For `x in (2,5], y darr` from `oo to 10`.

For x `in [5, oo), y uarr`, from `10 to oo`.

The complexity in rise and fall of y is understandable upon seeing

that the given equation has the form

`(y-x-2)(x-2)=9`.

This represents the hyperbola having asymptotes

( slant ) y = x +2 and ( vertical ) x = 2.

Respectively, there is rise and fall in the two branches.

See the illustrative graph.

graph{(y-x-2)(x-2)-9=0 [-80, 80, -40, 40]}

In my style, this is my answer. There ought to be some omissions or

additions, and corrections there upon, for improvement.

I request ( 1 other ) editors to give all that in comments, separately. It

is my duty to thank them, and edit my answer, accordingly..

Answer 2

The function is increasing when `x in ] -oo,-2 [ uu ]5, +oo[`

The function is decreasing when `x in] -1,2^(-) [ uu ]2^(+), 5[`

Explanation:

The domain of `f(x)` is `D_f(x)= RR-{2}`

We take the derivative of `f(x)` and when `f'(x)=0`, we obtain the critical points.

The derivative of a polynomial fraction is

`(u/v)'=(u'v-uv')/v^2`

Here, we have

`u=x^2+5`. `=>,`u'=2x#

`v=x-2`, =>`,`v'=1#

so,

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

`=(2x^2-4x-x^2-5)/(x-2)^2`

`=(x^2-4x-5)/(x-2)^2`

`=((x+1)(x-5))/(x-2)^2`

Therefore,

`f'(x)=0` , `=>`, `(x+1)(x-5)=0`

The critical points are `x=-1` and `x=5`

The denominator is `>0` for`AAx in D_f(x)`

Now we can form the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-1` `color(white)(aaaaa)` `2` `color(white)(aaaaaa)` `5` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+1` `color(white)(aaaa)` `-` `color(white)(aaa)` `0` `color(white)(a)` `+` `color(white)(a)` `∥` `color(white)(a)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `x-5` `color(white)(aaaa)` `-` `color(white)(aaa)` `0` `color(white)(a)` `-` `color(white)(a)` `∥` `color(white)(a)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaa)` `+` `color(white)(aaa)` `0` `color(white)(a)` `-` `color(white)(a)` `∥` `color(white)(a)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `↗` `color(white)(a)` `-2` `color(white)(a)` `↘` `color(white)()` `∥` `color(white)(a)` `↘` `color(white)(aa)` `10` `color(white)(aa)` `↗`

Therefore,

The function is increasing when `x in ] -oo,-2 [ uu ]5, +oo[`

The function is decreasing when `x in] -1,2^(-) [ uu ]2^(+), 5[`