Find increasing and decreasing intervals of a functionThe given function is f ( x ) = x 200 − x 100 , and I'm supposed to find it's decreasing and increasing intervals. Also, I should find them not by using derivatives but by doing function composition, like this: f 1 ( x ) = x 100 f 2 ( x ) = x ( x − 1 ) f ( x ) = f 2 ( f 1 ( x ) )I know that f 1 is decreasing on the interval ( − ∞ , 0 ] and increasing on [ 0 , + ∞ ), and f 2 is decreasing on ( − ∞ , 1 2 ] and increasing on [ 1 2 , + ∞ ), but I'm not really sure what to do next.

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Find increasing and decreasing intervals of a functionThe given function is   f  (  x  )  =      x          200        −      x          100      , and I&#039;m supposed to find it&#039;s decreasing and increasing intervals. Also, I should find them not by using derivatives but by doing function composition, like this:      f    1    (  x  )  =      x          100            f    2    (  x  )  =  x  (  x  −  1  )  f  (  x  )  =      f    2    (      f    1    (  x  )  )I know that       f    1   is decreasing on the interval   (  −  ∞  ,  0  ] and increasing on   [  0  ,  +  ∞  ), and       f    2   is decreasing on   (  −  ∞  ,      1    2    ] and increasing on   [      1    2    ,  +  ∞  ), but I&#039;m not really sure what to do next.

petyelebxu · Accepted Answer

Step 1First of all note that f(x) is an even function, i.e. symmetric in y-axis. Thus it is enough to discuss the behaviour of f(x) for   x  ≥  0. I use the same function       f    1    (  x  ) and       f    2    (  x  ) in the question.For   x  ≤  y  ≤            1              /            2        100   we have       f    1    (  x  )  ≤      f    1    (  y  )  ≤  1      /    2 so       f    2    (      f    1    (  x  )  )  ≥      f    2    (      f    1    (  y  )  ) hence   f  (  x  )  ≥  f  (  y  ). So f(x) is decreasing on   [  0  ,            1              /            2        100    ].Step 2For             1              /            2        100    ≤  x  ≤  y we have   1      /    2  ≤      f    1    (  x  )  ≤      f    1    (  y  ) so       f    2    (      f    1    (  x  )  )  ≤      f    2    (      f    1    (  y  )  ) hence   f  (  x  )  ≤  f  (  y  ). So (x)  is increasing on   [            1              /            2        100    ,  ∞  ).To keep the fun not only for myself, I let you do the work for   x  &amp;lt;  0

Fahdvfm · Accepted Answer

Step 1Once you take (as you did)       f    1    (  x  )  =      x          100       and       f    2    (  x  )  =      x    2    −  x, you can find the x coordinate of the vertex (here the minimum) of the parabola       f    2  , which is       x    0    =  −      b          2      a        =      1    2  , so we need       x          100        =      1    2      ⟹        x          1      ,      2        =  ±      1          2      100      .Then you can factorise       f    2    (      f    1    (  x  )  )  =      x          100            (          x              100              −    1    )    =      x          100            (          x              50              −    1    )        (          x              50              +    1    )    =      x          100            (          x              25              −    1    )    (      x          25        +  1  )  (      x          50        +  1  ).Roots of       f    2   are   x  ±  1 and   x  =  0.Now, investigate the behaviour of this product on the intervals   (  −  ∞  ,  −  1  ]  ,  (  −  1  ,  0  ]  ,  (  0  ,  1  ] and   (  1  ,  +  ∞  ).Step 2We can leave       x          100       and       x          50        +  1 aside as those are never negative, their restrictions to   (  −  ∞  ,  0  ) and   [  0  ,  +  ∞  ) are separately injective and there is an even number of such factors.                                −        ∞                          −        1                          0                          1                                        +                ∞                                                    x                      25                          −        1                                −                              |                            −                              |                            −                    ∘                    +                                      x                      25                          +        1                                −                    ∘                    +                              |                            +                              |                            +                            product                                +                          −                          −                          +            Our       x    0    =  ±      1          2      100       are exactly in the intervals   (  −  1  ,  0  ] and   (  0  ,  1  ] respectively.Now we have a better insight how   f  (  x  )  =      x          200        −      x          100       should look like       (    −    ∞    ,          1      2        ]    ∪      [          1      2        ,    +    ∞    )  . Then we examine the intervals       (    0    ,    +          1              2        100              ]   and       (          1              2        100              ,    +    ∞    )   to see that the overall composition is decreasing on       (    −    ∞    ,    −          1              2        100              )    ∪      (    0    ,          1              2        100              )   and increasing on       [    −          1              2        100              ,    0    ]    ∪      [          1              2        100              ,    +    ∞    )

The given function is f(x)=x^{200}-x^{100}, and I'm supposed to find it's decreasing and increasing intervals. Also, I should find them not by using derivatives but by doing function composition, like this: f_1(x)=x^{100}, f_2(x)=x(x-1), f(x)=f_2(f_1(x)).

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