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? Thanks!!
obviously n has to be even, because i need the signs of the limits of the quadratic $\displaystyle f(x)=2x^{n}-4x+1$ to be the same, when x tend to both +ve and -ve infinity. Then the graph of f(x) will curve either upwards or downwards at both ends, so there will exists some k where the equation has no solution. But how to give a formal solution??
or is there any other way to solve it???