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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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