Let f ( x ) = ( 3 x − 2 ) ( x 2 − 7 ). Find all the values of x for which f ′ ( x ) = 0

Question

Let   f  (  x  )  =  (  3  x  −  2  )  (      x    2    −  7  ). Find all the values of x for which       f    ′    (  x  )  =  0

Marcelo Stafford · Accepted Answer

Given:  f  (  x  )  =  (  3  x  −  2  )  (      x    2    −  7  )        f    ′    (  x  )  =      d          d      x        [  (  3  x  −  2  )  (      x    2    −  7  )  ]  [  ∵      d          d      x        u  v  =  u      v    ′    +  v      u    ′    ]    =  (  3  x  −  2  )      d          d      x        (      x    2    −  7  )  +  (      x    2    −  7  )      d          d      x        (  3  x  −  2  )    =  (  3  x  −  2  )  (  2  x  −  0  )  +  (      x    2    −  7  )  (  3  −  0  )    =  (  3  x  −  2  )  2  x  +  3  (      x    2    −  7  )    =  6      x    2    −  4  x  +  3      x    2    −  21    =  9      x    2    −  4  x  −  21      f    ′    (  x  )  =  0  ⇒  9      x    2    −  4  x  −  21  =  0    roots   =            −      b      ±                        b          2                −        4        a        c                    2      a          =            −      (      −      4      )      ±              (        −        4                  )          2                −        4        (        9        )        (        −        21        )                    2      (      9      )          =            4      ±              16        +        756              18    =            4      ±              772              18      =      4    8    ±            772        18      x  =      4    18    ±            193        9    =      2    9    ±            193        4      ∴  x  =      2    4    +            193        9    ,      2    9    −            193        9

Let f(x)=(3x-2)(x^2-7). Find all the values of x for which f'(x)=0

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