Determine the possible integer values of n for which there exists some real number k such that the equation $\displaystyle 2x^{n}-4x+1=k$ has no real roots.
anyone can help?
My guess is that n has to be even. My (informal) thinking goes like this.
If you think about the graph of the equation, you can see what happens when x goes to $\displaystyle \pm \infty$.
When n is odd, it starts at $\displaystyle -\infty$ when $\displaystyle x \to -\infty$ and goes to $\displaystyle +\infty$ when $\displaystyle x \to +\infty$. Somewhere it has to cross x=0 at least once.
However, if n is even then as $\displaystyle x \to -\infty$, we have that your equation goes to $\displaystyle +\infty$. It starts bigly positive and ends bigly positive and you can always assign a value of k to make sure it misses x=0 altogether.
You ought to be able to bag that up into a proof in whatever form that the level of your course requires.
my guess was n must be even, so let's take a look at what where the minimum/maximum is when n is even.
Take the derivative, set it to 0, plug it back in, and you just find a value for k that is less than that value, and factor it out.
though I'm not sure if that approach is useful. Maybe induction on n? and assume even so you say something like:
$\displaystyle 2x^{2n}-4x+1=k$ has no real roots.
and let that be your induction hypothesis?
For $\displaystyle n$ odd, as $\displaystyle x\to\infty, f(x)\to\infty$ and as $\displaystyle x\to-\infty, f(x)\to-\infty$. The Intermediate Value Theorem states that $\displaystyle f(x) = 0$ for some $\displaystyle x\in\mathbb{R}$.
So $\displaystyle n$ must be even.
Let $\displaystyle f(x)=2x^n-4x+1$. For $\displaystyle n$ even, we know that the minimum value of $\displaystyle f(x)$ occurs when $\displaystyle 2nx^{n-1}-4=0$, or when $\displaystyle x=\left(\frac{2}{n}\right)^{\frac{1}{n-1}}$. Call this point $\displaystyle x_0$. Therefore, if we let $\displaystyle k<-\left|f(x_0)\right|$, then $\displaystyle f(x)-k > 0 ~\forall~ x\in\mathbb{R}$. $\displaystyle \square$