'The gradient of a curve is given by $\displaystyle (dy)/(dx) = x^2 - 6x$. Find the set of values of x for which y is an increasing function of x.

Im pretty good at calculus, but i'm not sure what it means by an increasing function of x.

Any help?

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- Apr 24th 2008, 03:20 AMnugiboyCalculus question!
'The gradient of a curve is given by $\displaystyle (dy)/(dx) = x^2 - 6x$. Find the set of values of x for which y is an increasing function of x.

Im pretty good at calculus, but i'm not sure what it means by an increasing function of x.

Any help? - Apr 24th 2008, 03:22 AMtopsquark
- Apr 24th 2008, 03:28 AMnugiboy
- Apr 24th 2008, 04:10 AMIsomorphism
You want $\displaystyle x^2 - 6x > 0$

If x < 0, then x(x-6) > 0

If 0 < x < 6, then x(x-6) < 0

If x > 6, then x(x-6) > 0

So if $\displaystyle x \in ( - \infty, 0) \bigcup (6,\infty)$, then $\displaystyle \frac{dy}{dx}$ is positive.

Or in English, except for values of x between 0 and 6, everywhere it is increasing :D - Apr 24th 2008, 04:16 AMMoo
Hello,

A table of signs can help :

$\displaystyle x \leq 0 \text{ iff } x \leq 0$ (this can look unuseful, but it's the reasoning :))

$\displaystyle x-6 \leq 0 \text{ iff } x \leq 6$

So if $\displaystyle x \in ]-\infty,0] \ , \ x \leq 0 \text{ and } x-6 \leq 0. \text{ Hence } x(x-6) \geq 0$

If $\displaystyle x \in [0,6] \ , \ x \geq 0 \text{ and } x-6 \leq 0. \text{ Hence } x(x-6) \leq 0$

If $\displaystyle x \in [6,+\infty[ \ , \ x \geq 0 \text{ and } x-6 \geq 0. \text{ Hence } x(x-6) \geq 0$

So the function will be increasing for $\displaystyle \boxed{x \in ]- \infty,0] \cup [6,+\infty[} \ (= \mathbb{R} \backslash [0,6])$ :D