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Find the value of $a$ so the function $x^3-ax$ is increasing on $]-\infty, \infty[$?

So the derivative in $3x^2-a$. In order that the function always increases then $b^2-4ac \leq 0$ right? Which means $a \leq 0$, but the answer in my textbook is $a = 0$.

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    Took the liberty of editing title and tag.2012-12-30

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You are correct, the function is increasing for every choice of $a \le 0$.