Prove symmetry of natural logarithmProve 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?

Question

Prove symmetry of natural logarithmProve 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&#039;t know what to do? Is this the proof or did I miss something?

nuramaaji2000fh · Accepted Answer

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.

Aleah Booth · Accepted Answer

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 &quot;  ln&quot; is a one-one function:      x    2    =  (  2  a  −  x      )    2  Expanding and cancelling:  4      a    2    =  4  a  xIt gives two solutions:  a  =  0  ,  x  =  aSince 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.

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?

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