Show that the equation x 3 + 1 = 15 x has three solutions in the interval [ − 4 , 4 ]

Question

Show that the equation       x    3    +  1  =  15  x has three solutions in the interval   [  −  4  ,  4  ]

volksgeesr9 · Accepted Answer

Let   f  (  x  )  =      x    3    −  15  x  +  1; then   f          ′    (  x  )  =  3      x    2    −  15  =  3  (      x    2    −  5  )  =  3  (  x  −      5    )  (  x  +      5  . It’s easy to check that f has a local maximum at   x  =  −      5   and a local minimum at   x  =      5  , either by the second derivative test, by examining the sign of the first derivative, or simply by knowing the shape of the graph of a cubic polynomial.Now                    f        (        −        4        )                            =        −        64        +        60        +        1        =        −        3        &amp;lt;        0        ,                            f        (        −                  5                )                            =        −        5                  5                +        15                  5                +        1        =        10                  5                +        1        &amp;gt;        0        ,                            f        (                  5                )                            =        5                  5                −        15                  5                +        1        =        −        10                  5                +        1        &amp;lt;        0        ,         and                            f        (        4        )                            =        64        −        60        +        1        =        5        &amp;gt;        0                ,            so f(x) changes sign three times on the interval [−4,4] and by the intermediate value theorem must have (at least) three zeroes.Note that there are many ways to choose the test points, but using the critical points is an easy way to be sure of hitting good ones.

Show that the equation x^3+1=15x has three solutions in the interval [−4,4]

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New Questions in Calculus 1