For what values of $k$ is $g(x)=x^3+kx^2+x$ one-to-one? I need to find for what values of $k$ $g(x)=x^3+kx^2+x$ is one-to-one. I tried finding for what values it is strictly increasing and got the derivative to be $3x^2+2kx+1>0$, but I'm not really sure where to go from there since there are two variables.
 A: Hints:
Your function $g(x)$, being a cubic polynomial, is one-to-one if and only if the derivative has at most one root. One root is allowed.
You can tell how many roots a quadratic equation $ax^2+bx+c=0$ has by examining its discriminant $b^2-4ac$. If the discriminant is positive, there are two roots; if zero, one root; if negative, no real roots.
So examine the discriminant of the derivative of $g(x)$. That discriminant will depend on $k$, so find which values of $k$ will make the discriminant non-positive.
A: if the derivative had two roots, for $y$ between the local max and local min there would be three values of $x.$ you must require the discriminant $$k^2 - 3 \le 0$$ that means $$-\sqrt 3 \le k \le \sqrt 3. $$
A: You only need to find the value of one variable, $k$.
If you plot $y = 3x^2+2kx+1$ you get an upward-opening parabola.
To ensure that $y = 3x^2+2kx+1 > 0$ for all $x$ you need look only at the
minimum point on this parabola, since all others will have a greater value of $y$.
In general, the minimum of $ax^2 + bx + c$ ($a>0$) occurs at $x = -\frac{b}{2a}$.
If you apply that formula to your case ($a=3$ and $b=2k$), you will find the
minimum value of $3x^2+2kx+1$ as a function of $k$.
Now you just need to identify when this value is positive.
But a better alternative may be to look at the discriminant (as explained by abel and by Rory Daulton), since it requires less computation.
You should also consider carefully the case where $3x^2+2kx+1 \geq 0$ 
and equality is actually achieved at one value of $x$.
A function with a zero derivative may not be one-to-one, but some
functions that have zero derivatives for some values of $x$ can be one-to-one.
