Showing x 1 + x &lt; log ⁡ ( 1 + x ) &lt; x for all x &gt; 0 using the mean value theoremI want to show that x 1 + x &lt; log ⁡ ( 1 + x ) &lt; xfor all x &gt; 0 using the mean value theorem. I tried to prove the two inequalities separately. x 1 + x &lt; log ⁡ ( 1 + x ) ⇔ x 1 + x − log ⁡ ( 1 + x ) &lt; 0Let f ( x ) = x 1 + x − log ⁡ ( 1 + x ) .Since f ( 0 ) = 0and f ′ ( x ) = 1 ( 1 + x ) 2 − 1 1 + x &lt; 0for all x &gt; 0, f ( x ) &lt; 0 for all x &gt; 0. Is this correct so far?I go on with the second part: Let f ( x ) = log ⁡ ( x + 1 ). Choose a = 0 and x &gt; 0 so that there is, according to the mean value theorem, an x 0 between a and x with f ′ ( x 0 ) = f ( x ) − f ( a ) x − a ⇔ 1 x 0 + 1 = log ⁡ ( x + 1 ) x Since x 0 &gt; 0 ⇒ 1 x 0 + 1 &lt; 1. ⇒ 1 &gt; 1 x 0 + 1 = log ⁡ ( x + 1 ) x ⇒ x &gt; log ⁡ ( x + 1 )

Question

Showing       x          1      +      x        &amp;lt;  log  ⁡  (  1  +  x  )  &amp;lt;  x for all   x  &amp;gt;  0 using the mean value theoremI want to show that      x          1      +      x        &amp;lt;  log  ⁡  (  1  +  x  )  &amp;lt;  xfor all   x  &amp;gt;  0 using the mean value theorem. I tried to prove the two inequalities separately.      x          1      +      x        &amp;lt;  log  ⁡  (  1  +  x  )  ⇔      x          1      +      x        −  log  ⁡  (  1  +  x  )  &amp;lt;  0Let  f  (  x  )  =      x          1      +      x        −  log  ⁡  (  1  +  x  )  .Since  f  (  0  )  =  0and      f    ′    (  x  )  =      1          (      1      +      x              )        2              −      1          1      +      x        &amp;lt;  0for all   x  &amp;gt;  0,    f  (  x  )  &amp;lt;  0 for all   x  &amp;gt;  0. Is this correct so far?I go on with the second part: Let   f  (  x  )  =  log  ⁡  (  x  +  1  ). Choose   a  =  0 and   x  &amp;gt;  0 so that there is, according to the mean value theorem, an       x    0   between   a and   x with      f    ′    (      x    0    )  =            f      (      x      )      −      f      (      a      )              x      −      a        ⇔      1                  x        0            +      1        =            log      ⁡      (      x      +      1      )        x  Since      x    0    &amp;gt;  0  ⇒      1                  x        0            +      1        &amp;lt;  1.  ⇒  1  &amp;gt;      1                  x        0            +      1        =            log      ⁡      (      x      +      1      )        x    ⇒  x  &amp;gt;  log  ⁡  (  x  +  1  )

razdiralem · Accepted Answer

By Definition of log(which already a kind of mean value theorem ) we have For any   x  &amp;gt;  0 We have      x          x      +      1        =      ∫          0        x              d      t              x      +      1        ≤      ∫          0        x              d      t              t      +      1        =      ln    ⁡    (    x    +    1    )    =      ∫          0        x              d      t              t      +      1        ≤      ∫          0        x              d      t        1    =  xThus,      x          x      +      1        ≤  ln  ⁡  (  x  +  1  )  ≤  x

Lena Bell · Accepted Answer

Substitute   x  =  y  −  1 to get  1  −      y          −      1        &amp;lt;  log  ⁡  y  &amp;lt;  y  −  1for all   y  &amp;gt;  1. Now note      ∫    1    y        t          −      2        d  t  &amp;lt;      ∫    1    y        t          −      1        d  t  &amp;lt;      ∫    1    y    d  tfor   y  &amp;gt;  1

Showing x 1 + x </mrow> </mfrac> < log ⁡<!-- ⁡

Answered question

Answer & Explanation

New Questions in Algebra II