Finding all the set of values for k such that $-2x^3+24x+8=k$ 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?

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