find all the set of values for k such that $-2x^3+24x+8=k$ has more than one solution.

This problem can only be solved using basic Calculus 1 knowledge.

I thought about setting the equation to zero and then computing the x value that can then be rewritten in terms of k but I am not sure that is the correct approach can someone help me?

  • $\begingroup$ Hint: you're going to need to use the general solution to the cubic polynomial $x^3-12x-\frac{8-k}{2}=0$. (I suppose this isn't the Calc I solution though) $\endgroup$
    – Ian Coley
    Sep 18, 2013 at 3:59

3 Answers 3


Let $f(x) = -2x^3+24x+8 = k$

Now Let $f(x) = -2x^3+24x+8$ and we will find the nature of cure and plot the curve in $X-Y$ plane

So $f^{'}(x) = -6x^2+24$ and for Max. and Min. $f^{'}(x) = 0$

we get $-6(x^2-4) = 0 \Rightarrow x = \pm 2$

Now $f^{''}(x) = -12x$

So at $x = -2$ , we get $f^{''}(-2) = -12 \times -2 = 24>0$

So $x = -2$ is a point of Minimum

Similarly at $x = 2$ , we get $f^{''}(2) = -12 \times 2 = -24>0$

So $x = 2$ is a point of Maximum

So $f(-2) \leq f(x) \leq f(+2)$

So $-24 \leq f(x) \leq 40$

So value of $k$ for which the equation $f(x) = -2x^3+24x+8$ has more then one real roots is

$-24 < k < 40\Rightarrow k\in \left(-24,40\right)$

  • $\begingroup$ The inequalities are not strict. You want more than one solution, not necessarily more than two. $\endgroup$
    – Pipicito
    Sep 18, 2013 at 4:19

Hint: For the function to have more than one solution, it must have two distinct, real turning points. For what values of $k$ do those turning points occur on opposite sides of the $x$ axis?

(where we're taking the function to be $y=-2x^3+24x+8-k$)

  • $\begingroup$ Ok so I took the derivate and found $x=\sqrt{\frac{14}{3}}$. Is that correct and what do I do from here? Plug x back into the original solution? $\endgroup$ Sep 18, 2013 at 4:09
  • $\begingroup$ There should be two turning points, you've only identified one. From there, you need to work out when the function passes the $x$ axis more than once. I won't give you all of the details, as that would defeat the purpose of you having a go at the question. Try plotting the function, to see what it's doing, and think about the significance of my suggestion that you look for when the two turning points are on opposite sides of the $x$ axis. $\endgroup$
    – Glen O
    Sep 18, 2013 at 4:11

Define $f:\mathbb{R} \rightarrow \mathbb{R}$ with $f(x)= -2x^3+24x+8$. Study this function with the regular first derivative stuff considerations. You will have no problems with this as the derivative of $f$ is a second degree polynomial and you can then calculate ir zeros explicitly. Now sketch a graph. For which values of $k \in \mathbb{R}$ does the line $y=k$ has, at least, two points of intersection with the graph of $f$?

Note: the last part may not be totally formal but it can be modified to make it fulfill formality requirements. One possibility is to go like this: break the real axis into three intervals such that $f$ is injective in each one and then count the number of solutions in each piece in a formal way(exploiting the injectivity of $f$)


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