f(x)=x/(x^2+1) Find the absolute extreme values on [-7,7]

pancha3

pancha3

Answered question

2021-08-13

f(x)=xx2+1
Find the absolute extreme values on [-7,7]

Answer & Explanation

komunidadO

komunidadO

Skilled2021-08-14Added 86 answers

First direction:
f(x)=xx2+1 on [-7,7]
f(x)=ddx(xx2+1)
=(x2+1)x(2x)(x2+1)2
=1x2(x2+1)2
Equate above first derivative to zero to obtain the extreme points:
f(x)=0
1x2(x2+1)2=0
1x2=0
x2=1
x=±1
Compute the second derivative:
f(x)=1x2(x2+1)2
f(x)=ddx(1x2(x2+1)2)
=(x2+1)2(2x)(1x2)(4x(x2+1))(x2+1)4
f(1)=1(2)01=2<0 [max at 1]
f(1)=1(2)01=2<0 [min at -1]
Thus,
Max value is f(1)==of1(xx2+1)=12
Min value is f(1)==of1(xx2+1)=12

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