Find range of valuesFind the range of values of the constant a at which the equation x 3 − 3 a 2 x + 2 = 0 has 3 different real number roots.I took the derivative and found that x = − a , a. Then I solved for f ( a ) = 0 and f ( − a ) = 0 to find that a = − 1 , 1.How do I use this information to find the range of values, or am I on the wrong path completely?

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Find range of valuesFind the range of values of the constant a at which the equation       x    3    −  3      a    2    x  +  2  =  0 has 3 different real number roots.I took the derivative and found that   x  =  −  a  ,  a. Then I solved for   f  (  a  )  =  0 and   f  (  −  a  )  =  0 to find that   a  =  −  1  ,  1.How do I use this information to find the range of values, or am I on the wrong path completely?

Simon Zhang · Accepted Answer

Step 1You&#039;re on the wrong path. The points where       f    ′    =  0 are &quot;flat spots&quot; in the graph. That doesn&#039;t tell you anything about where it&#039;s zero.Actually, it tells you a little: between any two zeroes, there has to be a flat spot. So if   a  =  0, you can&#039;t possibly have three zeroes, because you&#039;ve got only one flat spot, at 0.Step 2On the other hand, you know that       f    ′    (  a  )  =  0 and       f    ′    (  −  a  )  =  0, and the graph&#039;s a cubic, i.e., it goes from lower left to upper right. If it happens that   f  (  −  a  )  &amp;gt;  0 and   f  (  a  )  &amp;lt;  0, then by the intermediate value theorem, f must have three zeroes.

Elias Heath · Accepted Answer

Step 1Since       f    ′    (  x  )  =  3      x    2    −  3      a    2    =  3  (  x  −  a  )  (  x  +  a  ) you see that f is increasing on   (  −  ∞  ,  −      |    a      |    ] and on   [      |    a      |    ,  +  ∞  ) and decreasing in   [  −      |    a      |    ,      |    a      |    ]. On every such interval you can have at most one solution to the equation   f  (  x  )  =  0, because of monotonicity. To have a solution in each interval you need the function to change sign on the extreme points of the interval.Step 2So to have three solutions you need   f  (  −      |    a      |    )  &amp;gt;  0 and   f  (      |    a      |    )  &amp;lt;  0 which gives:  2      |    a            |        3    +  2  &amp;gt;  0    −  2      |    a            |        3    +  2  &amp;lt;  0which is true for       |    a      |    &amp;gt;  1.

Find the range of values of the constant a at which the equation x^3-3a^2x+2=0 has 3 different real number roots.

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