Interval of monotonicity of the function f ( x ) = ln ⁡ ( π + x ) ln ⁡ ( e + x ) .Find the exact interval in which the function f ( x ) = ln ⁡ ( π + x ) ln ⁡ ( e + x ) is increasing and decreasing.My try: The domain of the function is ( − e , ∞ )The derivative is: f ′ ( x ) = ( e + x ) ln ⁡ ( e + x ) − ( π + x ) ln ⁡ ( π + x ) ( π + x ) ( e + x ) ln 2 ⁡ ( e + x ) .Since due to the domain we have e + x &gt; 0 , π + x &gt; 0 and trivially ln 2 ⁡ ( e + x ) &gt; 0.Now it is sufficient to check the nature of numerator above.Let p = e + x , q = π + x.The numerator is in the form p ln ⁡ ( p ) − q ln ⁡ ( q ) where p &lt; q , ∀ x.Now we know that the function h ( x ) = x ln ⁡ x is increasing in [ 1 e , ∞ ) .Case 1. If 1 e ≤ p &lt; q we have: h ( p ) &lt; h ( q ) ⟹ p ln ⁡ p − q ln ⁡ q &lt; 0Thus f ′ ( x ) &lt; 0 , ∀ x ∈ [ 1 e − e , ∞ ) = [ − 2.35 , ∞ )So f(x) must be decreasing in the above interval. But i compared f(-2) and f(-1) It is the other way round, that is f ( − 2 ) &lt; f ( − 1 )Now what i did is i used graphing calculator to check the graph. Yes it is decreasing, but there is a vertical asymptote. How to identify the equation of vertical asymptote?Also the graph is decreasing in [ − 2.541 , ∞ ) which is not matching with my answer above.

Question

Interval of monotonicity of the function   f  (  x  )  =            ln      ⁡      (      π      +      x      )              ln      ⁡      (      e      +      x      )      .Find the exact interval in which the function   f  (  x  )  =            ln      ⁡      (      π      +      x      )              ln      ⁡      (      e      +      x      )       is increasing and decreasing.My try: The domain of the function is   (  −  e  ,  ∞  )The derivative is:       f    ′    (  x  )  =            (      e      +      x      )      ln      ⁡      (      e      +      x      )      −      (      π      +      x      )      ln      ⁡      (      π      +      x      )              (      π      +      x      )      (      e      +      x      )              ln        2            ⁡      (      e      +      x      )      .Since due to the domain we have   e  +  x  &amp;gt;  0  ,  π  +  x  &amp;gt;  0 and trivially       ln    2    ⁡  (  e  +  x  )  &amp;gt;  0.Now it is sufficient to check the nature of numerator above.Let   p  =  e  +  x  ,  q  =  π  +  x.The numerator is in the form   p  ln  ⁡  (  p  )  −  q  ln  ⁡  (  q  ) where   p  &amp;lt;  q  ,  ∀  x.Now we know that the function   h  (  x  )  =  x  ln  ⁡  x is increasing in       [          1      e        ,    ∞    )  .Case 1. If       1    e    ≤  p  &amp;lt;  q we have:  h  (  p  )  &amp;lt;  h  (  q  )    ⟹    p  ln  ⁡  p  −  q  ln  ⁡  q  &amp;lt;  0Thus       f    ′    (  x  )  &amp;lt;  0  ,  ∀  x  ∈      [          1      e        −    e    ,    ∞    )    =  [  −  2.35  ,  ∞  )So f(x) must be decreasing in the above interval. But i compared f(-2) and f(-1) It is the other way round, that is    f  (  −  2  )  &amp;lt;  f  (  −  1  )Now what i did is i used graphing calculator to check the graph. Yes it is decreasing, but there is a vertical asymptote. How to identify the equation of vertical asymptote?Also the graph is decreasing in   [  −  2.541  ,  ∞  ) which is not matching with my answer above.

Kenley Rasmussen · Accepted Answer

Step 1The domain consists of those x for which   π  +  x  &amp;gt;  0  ,  e  +  x  &amp;gt;  0 and   ln  ⁡  (  e  +  x  )  ≠  0. That gives the set   (  −  e  ,  1  −  e  )  ∪  (  1  −  e  ,  ∞  ). Furthermore,       lim          x      →      (      1      −      e              )        −              f  (  x  )  =  −  ∞ and       lim          x      →      (      1      −      e              )        +              f  (  x  )  =  ∞ so there&#039;s a vertical asymptote   x  =  1  −  e. Next, the problem involves determining the sign of   g  (  x  )  =  (  e  +  x  )  ln  ⁡  (  e  +  x  )  −  (  π  +  x  )  ln  ⁡  (  π  +  x  )   .Step 2I don&#039;t think the answer here can be given in closed form. What you can do is show that g is strictly decreasing (by checking the derivative) and since       lim          x      →      (      −      e              )        +              g  (  x  )  =  −  (  π  −  e  )  ln  ⁡  (  π  −  e  )  &amp;gt;  0 and   g  (  −  2  )  =  (  e  −  2  )  ln  ⁡  (  e  −  2  )  −  (  π  −  2  )  ln  ⁡  (  π  −  2  )  &amp;lt;  0, g has a unique zero   α  ∈  (  −  e  ,  −  2  ). Thus,   g  (  x  )  &amp;gt;  0 for   x  ∈  (  −  e  ,  α  ) and   g  (  x  )  &amp;lt;  0 for   (  α  ,  ∞  ). This implies that f is increasing in   (  −  e  ,  α  ] and decreasing in   [  α  ,  1  −  e  ) and   (  1  −  e  ,  ∞  ).Note, however, that f is not decreasing in   [  α  ,  1  −  e  )  ∪  (  1  −  e  ,  ∞  ). This is due to the vertical asymptote   x  =  1  −  e. So finding that asymptote was important.

Find the exact interval in which the function f(x)=ln(pi+x)/ln(e+x) is increasing and decreasing.

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