Prove that $a\geq0$, when $f(x)=x^{3}+ax$ is (strictly) increasing.

Question

I need help with proving that $a\geq0$, when $f(x)=x^{3}+ax$ is (strictly) increasing.

Edit: Without using derivatives.

Answer 1

Hint: Consider the derivative of $f$.

Answer 2

If a function $f(x)$ is strictly increasing then $f'(x) > 0 \forall x$

$$f'(x) = 3x^2 + a$$

Now assuming you meant $a>0$ we know that $f'(x) >0$ for all $x$ as the smallest $x^2$ can be is $0$ and adding $a$ makes $f'(x)>0 \forall x$

Answer 3

Prove that $$f(x_1)>f(x_2)\iff (x_1^3-x_2^3)+a(x_1-x_2)>0$$ where $x_1>x_2$ and $a\geq 0$ and if $a<0$ you have that $f(x)=0$ when $x=0$ or $x=\sqrt{-a}$ which violates the strictness,also since $\frac{\sqrt{-a}}{2}>0$ you could also take $f(\frac{\sqrt{-a}}{2})=\frac{7a\sqrt{-a}}{8}<0$ and $f(0)=0$

Answer 4

Hint for the revised question ("without using derivatives").

Essentially, you can mimic what the derivative would tell you for values of $x$ near $0$. If $a$ were negative then $f(x) = x^2(x+a)$ would be decreasing for some range of values of $x$ you should be able to figure out.

Answer 5

The function $x^3+ax$ is odd. Therefore, if it is strictly increasing, it must take values in the first and third quadrant only.

Write $x^3+ax = x (x^2+a)$ as a product of two terms and consider its sign for $x > 0$. If $a < 0$, $x^2+a < 0$ for $|x|< \sqrt{-a}$ and thus $x (x^2+a) < 0$ for $0<x<\sqrt{-a}$, which contradicts the above. Hence $a \ge 0$.