For what real values of $k$ does $2x^3-3x^{21}-2x+1=k$ have exactly one real root?

Question

This is meant to be a non-calculator university interview-style question.

From graphing software of the function on the LHS I can see the solution is

$k<0.23$, $0.43<k<1.57$, $k>1.77$

and due to the symmetry around $y=1$, only the turning points at $y=0.23$ and $y=0.43$ need to be found.

The problem I have is in actually finding these turning points "by hand". Differentiating leads to a polynomial of degree $20$ with no straightforward solution.

Perhaps someone can point me in the right direction.

Answer 1

Assuming $2x^3-3x^2-2x+1=k$ instead of $2x^3-3x^{21}-2x+1=k$,

Let $f(x)=2x^3-3x^2-2x+1$

Then $f'(x)=6x^2-6x-2$

Thus, we can find the roots $\alpha$ and $\beta$ of $f'(x)=0$

Using the nature of the cubic's plot (not done using a graphing calculator), it is clear that only one solution exists if $k>f(\alpha)$ or $k<f(\beta)$, where $\alpha$ is the smaller root.

Answer 2

If you recall the procedure for solving the cubic equation $$2x^3-3x^2-2x+(1-k)=0$$ you firs eliminate the $x^2$ term to face a depressed cubic.

$$x=t- \frac{-3}{3\times 2}=t+ \frac 1{2}\quad \implies t^3-\frac{7 }{4}t-\frac{2k+1}{4} =0$$

Now, the dicriminant $$\Delta=-4 \left(-\frac 74\right)^3-27\left(-\frac{2k+1}{4} \right)^2=-\frac 14 \left(27 k^2+27 k-79\right)$$ must be negative. So, any $k$ outside the roots of this quadratic will lead to a single real root.