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

##### 2 Answers
Dec 22, 2016

See the explanation.

#### Explanation:

By actual division,

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

$y ' = 1 - \frac{9}{x - 2} ^ 2 = 0$, when ${\left(x - 2\right)}^{2} = 9 \to x = - 1 \mathmr{and} 5.$

For $x \in \left(- \infty , - 1\right] , y \uparrow ,$ from $- \infty \to - 2$..

For $x \in \left[- 1 , 2\right) , y \downarrow \to - \infty$.

For $x \in \left(2 , 5\right] , y \downarrow$ from $\infty \to 10$.

For x $\in \left[5 , \infty\right) , y \uparrow$, from $10 \to \infty$.

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

that the given equation has the form

$\left(y - x - 2\right) \left(x - 2\right) = 9$.

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..

Dec 22, 2016

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 $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{2\right\}$

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

The derivative of a polynomial fraction is

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{v} ^ 2$

Here, we have

$u = {x}^{2} + 5$. $\implies ,$u'=2x

$v = x - 2$, =>$,$v'=1

so,

$f ' \left(x\right) = \frac{2 x \left(x - 2\right) - 1 \left({x}^{2} + 5\right)}{x - 2} ^ 2$

$= \frac{2 {x}^{2} - 4 x - {x}^{2} - 5}{x - 2} ^ 2$

$= \frac{{x}^{2} - 4 x - 5}{x - 2} ^ 2$

$= \frac{\left(x + 1\right) \left(x - 5\right)}{x - 2} ^ 2$

Therefore,

$f ' \left(x\right) = 0$ , $\implies$, $\left(x + 1\right) \left(x - 5\right) = 0$

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

The denominator is $> 0$ for$\forall x \in {D}_{f} \left(x\right)$

Now we can form the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a}$$2$$\textcolor{w h i t e}{a a a a a a}$$5$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 1$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a}$∥$\textcolor{w h i t e}{a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 5$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a}$∥$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a}$∥$\textcolor{w h i t e}{a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$↗$\textcolor{w h i t e}{a}$$- 2$$\textcolor{w h i t e}{a}$↘$\textcolor{w h i t e}{}$∥$\textcolor{w h i t e}{a}$↘$\textcolor{w h i t e}{a a}$$10$$\textcolor{w h i t e}{a a}$↗

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[