Calculus application of derivativesLet g ( x ) = 2 f ( x / 2 ) + f ( 2 − x ) and f ″ ( x ) &lt; 0 for all x ∈ ( 0 , 2 ). If g(x) increases in (a,b) and decreases in (c,d) find the values of a, b, c and d.What I thought was a little graphical approach. I figured it out that if f′′(x) is less than zero then f′(x) would be decreasing (not considering the concavity) and the behaviour of g(x) would depend upon f(x). But I couldn't figure out the exact intervals.

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Calculus application of derivativesLet   g  (  x  )  =  2  f  (  x      /    2  )  +  f  (  2  −  x  ) and       f    ″    (  x  )  &amp;lt;  0 for all   x  ∈  (  0  ,  2  ). If g(x)  increases in (a,b) and decreases in (c,d) find the values of a, b, c and d.What I thought was a little graphical approach. I figured it out that if f′′(x) is less than zero then f′(x) would be decreasing (not considering the concavity) and the behaviour of g(x) would depend upon f(x). But I couldn&#039;t figure out the exact intervals.

ehedem26 · Accepted Answer

Step 1I figured it out that if f′′(x) is less than zero then f′(x) would be decreasingIt&#039;s enough. Look:      g    ′    (  x  )  =  2      f    ′    (  x      /    2  )  ⋅      1    2    +  f  (  2  −  x  )  ⋅  (  −  1  )  =      f    ′    (  x      /    2  )  −  f  (  2  −  x  )  .Step 21.   0  &amp;lt;  x      /    2  &amp;lt;  2  −  x; since f′(x) is decreasing then       g    ′    (  x  )  &amp;gt;  0 and g(x) is increasing.2.   x      /    2  &amp;gt;  2  −  x  &amp;gt;  0; since f′(x) is decreasing then       g    ′    (  x  )  &amp;lt;  0 and g(x) is decreasing.I think you can find a, b, c, d now.

Cyrus Travis · Accepted Answer

Step 1                    g        (        x        )                            =        2        f                  (                      x            2                    )                +        f        (        2        −        x        )                                      g          ′                (        x        )                            =                  f          ′                          (                      x            2                    )                −                  f          ′                (        2        −        x        )                                      g          ′                          (                      4            3                    )                                    =                  f          ′                          (                      2            3                    )                −                  f          ′                          (                      2            3                    )                                                  =        0            so             4      3       is a critical point for g(x). (I found that critical point by solving             x      2        =  2  −  x.)       f    ″    (  x  )  &amp;lt;  0 implies that f′(x) is decreasing, so  u  &amp;lt;      2    3      ⟹        f    ′    (  u  )  &amp;gt;      f    ′        (          2      3        )    u  &amp;gt;      2    3      ⟹        f    ′    (  u  )  &amp;lt;      f    ′        (          2      3        )  Step 2Thus, for   x  &amp;lt;            4      3      ,             x      2        &amp;lt;            2      3       and   2  −  x  &amp;gt;            2      3      , so                              g          ′                (        x        )                            =                  f          ′                          (                      x            2                    )                −                  f          ′                (        2        −        x        )                                          &amp;gt;                  f          ′                          (                      2            3                    )                −                  f          ′                          (                      2            3                    )                                                  =        0            and similarly   x  &amp;gt;            4      3          ⟹        g    ′    (  x  )  &amp;lt;  0. Thus g(x) increases for   x  &amp;lt;            4      3       and increases for   x  &amp;gt;            4      3      .Now we know the behavior of f only for   0  &amp;lt;  x  &amp;lt;  2. The function g is then known only for the intersection of   0  &amp;lt;            x      2        &amp;lt;  2 and   0  &amp;lt;  2  −  x  &amp;lt;  2, which is again   0  &amp;lt;  x  &amp;lt;  2.Therefore g(x) increases in       (    0    ,                  4        3              )   and decreases in       (                  4        3              ,    2    )   and we don&#039;t know what happens outside the interval (0,2).So   a  =  0  ,  b  =  c  =            4      3        ,  d  =  2.

Let g(x)=2f(x/2)+f(2-x) and f′′(x)<0 for all x in (0,2). If g(x) increases in (a,b) and decreases in (c,d) find the values of a, b, c and d.

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