Finding the number of solutions of an equation

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Homework Statement


Let hk : R -> R whereh_{k}(x)=x^3-6x+k and k is a real number

Find the value(s) of k for which the equation hk has 1,2 or 3 solutions.

The Attempt at a Solution



Don't know how to approach this.

Thanks!
 
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Can you graph the function?
Find max min and do that. It should help.
 
Hi,

Yes, i have graphed the function with the value of k being -8,0,6 and 10 (that was a previous question), but the value k can be anything for this question. How can I find the number of solutions produced?
 
How many times does the graph cross the x-axis? That's the number of solutions.

Obviously, if k= 0 then the function (not equation) is h_0(x)= x^3- 6x and presumably you are talking about the equation h_0(x)= x^3- 6x= x(x^2- 6)= 0 which has 3 solutions, 0 and \pm\sqrt{6}. Now try a few other values of k and see how the graph changes and how many times the graph crosses the x-axis.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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