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Grade 11Integral Calculus

Find the intervals in which f (x) =
|x−2|/
(x2)
is strictly increasing and strictly decreasing.

Profile image of KOMATI UMESH
12 Years agoGrade 11
Answers icon

3 Answers

Profile image of Parvez ali
ApprovedApproved Tutor Answer12 Years ago
f(x)= (x-2)/x^2 if x>=2
(2-x)/x^2 if x<2
= 1/x-2/x^2 if x>=2
2/x^2-1/x if x<2
df(x)/dx= -1/x^2+4/x^3 if x>=2
= 1/x^2-4/x^3 if x<2
now, (4-x)/x^3>0 => x(x-4)<0 => 0<x<4
df/dx>0 => 2=<x<4 and x<2=> increasing for x<4
df/dx<0=> x>4=> decreasing for x>4
parvez ali
askiitian faculty
b-tech
ism dhanbad
Profile image of Ajay Verma
ApprovedApproved Tutor Answer12 Years ago
Ans:

f(x)= |x-2|/ x2

case 1:
x-2 < 0 or x<2 ............................................. (1)

then f(X) = - (x-2)/ x2
= -1/x + 2/x2
for strictly increasing d f(x)/ dx >0
d f(x)/ dx = 1/x2 - 4/x3 > 0
1/x2 ( 1- 4/x) >0
1/x2 is always >0 so we take.. ( 1- 4/x) >0
or 1 > 4/x
sub_case_1 if x>0 then x >4................................ (2)
sub_case_2 if x< 0 then x<4.................................(3)
solving eqn1 , eqn2, eqn3.. we get for x<0 ....df(x)/dx >0 so strictly increasing

for strictly decreasing : df(x)/dx <0
d f(x)/ dx = 1/x2- 4/x3 < 0
1/x2 ( 1- 4/x) <0
1/x2is always >0 so.. ( 1- 4/x) <0
or 1 < 4/x
sub_case_3 if x>0 then x <4................................ (4)
sub_case_4 if x< 0 then x>4.......................... ........(5)

solving eqn1 , eqn4, eqn5 we get.. for 0< x < 2 df(x)/dx <0 so strictly decreasing



case 2:
x-2>0 or x>2 ..................................... (6)

then f(X) = (x-2)/ x2
= 1/x - 2/x2
for strictly increasing d f(x)/ dx >0
d f(x)/ dx = -1/x2+ 4/x3 > 0
1/x2 ( -1 +4/x) >0
1/x2is always >0 so we take.. ( -1+ 4/x) >0
or 1 < 4/x
sub_case_1 if x>0 then x <4................................ (7)
sub_case_2 if x< 0 then x>4.................................(8)

solving eqn6 , eqn7, eqn8 we get..for 2<x<4 df(x)/dx >0 so strictly increasing..

for decreasing d f(x)/ dx <0
d f(x)/ dx = -1/x2+ 4/x3 < 0
1/x2 ( -1 +4/x) <0
1/x2is always >0 so we take.. ( -1+ 4/x) <0
or 1 > 4/x
sub_case_1 if x>0 then x >4................................ (9)
sub_case_2 if x< 0 then x<4.................................(10)

solving eqn6 , eqn9, eqn10 we get..for x < 4 df(x)/dx <0 so strictly decreasing

so (0,2)U (4, infinty) decreasing
( - infinity, 0) U (2,4) increasing



Thanks and Regards,
Ajay verma,
askIITians faculty,
IIT HYDERABAD
Profile image of KOMATI UMESH
12 Years ago
THANKYOU

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