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Find the intervals in which f(x) = jx??2j x2 is strictly increasing and strictly decreasing.

Find the intervals in which f(x) = jx??2j
x2 is strictly increasing and strictly decreasing.

Grade:upto college level

4 Answers

bharat bajaj IIT Delhi
askIITians Faculty 122 Points
7 years ago
Please post the questain again. We could not understand your question.
Thanks
Bharat Bajaj
askIITians Faculty
Qualification.
IIT Delhi
sudhir pal
askIITians Faculty 26 Points
7 years ago
your expression is not so clear so I take the liberty to interpret it in my way
f(x) = |x2|x2

Since x2 is always greater than 0 so mod of x2 is always positive

So f(x) becomes x4

which is decreasing when x<0 and increasing when x>0




Thanks & Regards
Sudhir,
askIITians Faculty
Qualification.
IIT Delhi

Sher Mohammad IIT Delhi
askIITians Faculty 174 Points
7 years ago
175-1573_Untitled.jpg

Sher Mohammad,
IIT Delhi, B.Tech
Sher Mohammad IIT Delhi
askIITians Faculty 174 Points
7 years ago
f(x)=|x-2|/x^2 for x>2
=2-x/x^2 for x<2
f'(x)=(4/x^3)-(1/x^2) for x>2
=-((4/x^3)-(1/x^2)) for x<2
for strictly increasing f'(x)>0
(4/x^3)-(1/x^2) >0 for x>2
x>4 for x>2 imply 2<x<4
-((4/x^3)-(1/x^2)) >0 for x<2
1/x^2<-4/x^3 for x<0 imply x<-4
for strictly decreasing f'(x)<0
(4/x^3)-(1/x^2) <0 for x>2
x<4 for x>2 imply x>4
-((4/x^3)-(1/x^2)) <0 for x<2
1/x^2<-4/x^3 for x<2 imply -4<x<2

Sher Mohammad
B.Tech, IIT Delhi.

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