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

Answer:

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..

Dec 22, 2016

Answer:

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[ #