What are the points of inflection, if any, of f ( x ) = 5 cos 2 x − 10 sin x on x ∈ [ 0 , 2 π ] ?

Question

What are the points of inflection, if any, of             f              (        x        )              =    5                  cos        2            x        −    10          sin      x       on       x    ∈          [      0      ,      2      π      ]      ?

Spencer Shepard · Accepted Answer

We have:             f              (        x        )              =    5                  cos                  2                            (        x        )              -    10          sin              (        x        )            ;       x    ∈          [      0      ,      2      π      ]      To find the points of inflection, we must calculate the function&#039;s second derivative:      ⇒    f    ′          (      x      )        =                  d                              d          x                            (      5                        cos                      2                                    (          x          )                    )        -                  d                              d          x                            (      10              sin                  (          x          )                    )            ⇒    f    ′          (      x      )        =          (      5      ⋅      -              sin                  (          x          )                    ⋅      2              cos                  (          x          )                    )        -          (      10      ⋅              cos                  (          x          )                    )            ⇒    f    ′          (      x      )        =    -    10          sin              (        x        )                    cos              (        x        )              -    10          cos              (        x        )                  ⇒    f    ′    ′          (      x      )        =                  d                              d          x                            (      -      10              sin                  (          x          )                            cos                  (          x          )                    -      10              cos                  (          x          )                    )            ⇒    f    ′    ′          (      x      )        =          (      -      10      ⋅              (                  sin                      (            x            )                          ⋅        -                  sin                      (            x            )                          +                  cos                      (            x            )                          ⋅                  cos                      (            x            )                          )            )        -          (      10      ⋅      -              sin                  (          x          )                    )            ⇒    f    ′    ′          (      x      )        =    -    10          (      -                        sin                      2                                    (          x          )                    +                        cos                      2                                    (          x          )                    )        +    10          sin              (        x        )                  ⇒    f    ′    ′          (      x      )        =    -    10          (                        cos                      2                                    (          x          )                    -                        sin                      2                                    (          x          )                    )        +    10          sin              (        x        )            The points of inflection are those where the second derivative equals zero:      ⇒    f    ′    ′          (      x      )        =    0        ⇒    -    10          (                        cos                      2                                    (          x          )                    -                        sin                      2                                    (          x          )                    )        +    10          sin              (        x        )              =    0  One of the Pythagorean identities is                     cos                  2                            (        x        )              +                  sin                  2                            (        x        )              =    1  .We can rearrange this so that we get:      ⇒                  cos                  2                            (        x        )              =    1    -                  sin                  2                            (        x        )            Let&#039;s apply this rearranged identity to get:      ⇒    -    10          (              (        1        -                              sin                          2                                            (            x            )                          )            -                        sin                      2                                    (          x          )                    )        +    10          sin              (        x        )              =    0        ⇒    -    10          (      1      -      2                        sin                      2                                    (          x          )                    )        +    10          sin              (        x        )              =    0        ⇒    -    10    +    20                  sin                  2                            (        x        )              +    10          sin              (        x        )              =    0  Rearrange this to create a quadratic equation:      ⇒    20                  sin                  2                            (        x        )              +    10          sin              (        x        )              -    10    =    0        ⇒    10          (      2                        sin                      2                                    (          x          )                    +              sin                  (          x          )                    -      1      )        =    0        ⇒    2                  sin                  2                            (        x        )              +          sin              (        x        )              -    1    =    0  Thus, let&#039;s factorise using the middle-term break:      ⇒    2                  sin                  2                            (        x        )              +    2          sin              (        x        )              -          sin              (        x        )              -    1    =    0        ⇒    2          sin              (        x        )                    (              sin                  (          x          )                    +      1      )        -    1          (              sin                  (          x          )                    +      1      )        =    0        ⇒          (              sin                  (          x          )                    +      1      )              (      2              sin                  (          x          )                    -      1      )        =    0  We now have a product equal to zero, so one of the multiples must also be equal to zero:      ⇒          sin              (        x        )              +    1    =    0        ⇒          sin              (        x        )              =    -    1        ⇒    x    =          arcsin              (        -        1        )                  ⇒    x    =                  3        π                    2              +    2    π    n  or      ⇒    2          sin              (        x        )              -    1    =    0        ⇒    2          sin              (        x        )              =    1        ⇒          sin              (        x        )              =                  1                    2                  ⇒    x    =          arcsin              (                              1                                2                          )                  ⇒    x    =                  π                    6              +    2    π    n    ,                  5        π                    6              +    2    π    n  However, the domain is given as       x    ∈          [      0      ,      2      π      ]      .Hence, the points of inflection occur at       x    =                  π                    6            ,       x    =                  5        π                    6             and       x    =                  3        π                    2            .

What are the points of inflection, if any, of f(x)=5 cos^2 x-10 sin x on x in [0, 2pi]?

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