find the minimum value of x^2+1/x^2 when x is a real number.
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Answer:
Let z=
x
2
+x+1
x
2
−x+1
⇒z=
x
2
+x+1
x
2
+x+1−2x
⇒z=
x
2
+x+1
x
2
+x+1
−
x
2
+x+1
2x
⇒z=1−
x
2
+x+1
2x
Let y=
x
2
+x+1
2x
dx
dy
=
(x
2
+x+1)
2
(x
2
+x+1)×2−(2x)(2x+1)
dx
dy
=
(x
2
+x+1)
2
2(x
2
+x+1−x(2x+1))
dx
dy
=
(x
2
+x+1)
2
2(x
2
+x+1−2x
2
−x)
dx
dy
=
(x
2
+x+1)
2
2(−x
2
+1)
Maximum value occurs when
dx
dy
=0
⇒
(x
2
+x+1)
2
2(−x
2
+1)
=0
⇒−x
2
+1=0
⇒x
2
−1=0
⇒(x−1)(x+1)=0
∴x=−1,1
For x=−1,y=
x
2
+x+1
2x
=
1−1+1
−2
=−2
For x=1,y=
x
2
+x+1
2x
=
1+1+1
2
=
3
2
∴z
min
=1−y
max
=1+2=3 for y=−2
and z
min
=1−y
max
=1−
3
2
=
3
1
for y=
3
2
Since
3
1
<3
Thus, the least value is
3
1
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