Prove that f(x)=ln sqrt(x^2+1) is symmetrical in x=0. ln sqrt((x-a)^2+1)=ln sqrt((x+a)^2+1) sqrt((x-a)^2+1)=sqrt((x+a)^2+1) (x−a)^2+1=(x+a)^2+1 x^2−2ax+a^2+1=x^2+2ax+a^2+1 −2ax=2ax −x=x? I don't know what to do? Is this the proof or did I miss something?

aanpalendmw 2022-07-16 Answered
Prove symmetry of natural logarithm
Prove that f ( x ) = ln x 2 + 1 is symmetrical in x = 0
ln ( x a ) 2 + 1 = ln ( x + a ) 2 + 1
( x a ) 2 + 1 = ( x + a ) 2 + 1
( x a ) 2 + 1 = ( x + a ) 2 + 1
x 2 2 a x + a 2 + 1 = x 2 + 2 a x + a 2 + 1
2 a x = 2 a x
2 a x = 2 a x
x = x?
I don't know what to do? Is this the proof or did I miss something?
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Answers (2)

nuramaaji2000fh
Answered 2022-07-17 Author has 18 answers
If you can prove that your function is an even function that is
f ( x ) = f ( x )
then it is symmetrical about the y-axis or the line x = 0. In your case we have
f ( x ) = ln ( ( x ) 2 + 1 ) = ln ( ( x ) 2 + 1 ) = f ( x )
which proves the symmetry.
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Aleah Booth
Answered 2022-07-18 Author has 5 answers
If a function is symmetrical about a line x = a then it can be said that:
f ( x ) = f ( 2 a x ) x R
ln x 2 + 1 = ln ( 2 a x ) 2 + 1
By using property ln a b = b ln a
ln ( x 2 + 1 ) = ln ( ( 2 a x ) 2 + 1 )
Since " ln" is a one-one function:
x 2 = ( 2 a x ) 2
Expanding and cancelling:
4 a 2 = 4 a x
It gives two solutions:
a = 0 , x = a
Since we want symmetry for all real points we take the solution line x = a ( = 0 )
For this question there was an easy way: You could just have shown that
x 2 = ( x ) 2
For this question there was an easy way: You could just have shown that
x 2 = ( x ) 2
and the rest follows after.
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