How to use the second derivative test to find all relative extrema of f ( x ) = 5 + 3 x 2 - x 3 ?

Question

How to use the second derivative test to find all relative extrema of             f              (        x        )              =    5    +    3          x      2        -          x      3      ?

Vincent Burke · Accepted Answer

The first derivative is       f    ′          (      x      )        =    6    x    -    3          x      2        =    3    x          (      2      -      x      )      , which has roots at       x    =    0   and       x    =    2  .  These are the turning point as well as the potential sites of local extrema.Since the second derivative is       f    ′    ′          (      x      )        =    6    -    6    x  , we get       f    ′    ′          (      0      )        =    6    &amp;gt;    0   and       f    ′    ′          (      2      )        =    -    6    &amp;lt;    0  .  The fact that       f    ′    ′          (      0      )        &amp;gt;    0   (and the fact that       f    ′    ′   is continuous) implies that the graph of       f   is concave up near       x    =    0  , making, by the Second Derivative Test,        x    =    0   the location of a local minimum.  The fact that       f    ′    ′          (      2      )        &amp;lt;    0   (and the fact that       f    ′    ′   is continuous) implies that the graph of       f   is concave down near       x    =    2  , making, by the Second Derivative Test,        x    =    2   the location of a local maximum.  The local minimum value (output) is             f              (        0        )              =    5   and the local maximum value (output) is             f              (        2        )              =    5    +    12    -    8    =    9  .

How to use the second derivative test to find all relative extrema of f(x)=5+3x^2-x^3?

Answered question

Answer & Explanation

New Questions in Calculus 1