1. ## Calculus 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?

2. Originally Posted by nugiboy
'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?
What does the derivative of a function mean? Thus if it is positive/negative what does that mean about the function?

-Dan

3. Originally Posted by topsquark
What does the derivative of a function mean? Thus if it is positive/negative what does that mean about the function?

-Dan
Ah right so i need to find the values of x for which the gradient is positive?
Do i just do this by trial and error substituting in x values, or is there something im missing?

4. Originally Posted by nugiboy
Ah right so i need to find the values of x for which the gradient is positive?
Do i just do this by trial and error substituting in x values, or is there something im missing?
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

5. 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 $\displaystyle \boxed{x \in ]- \infty,0] \cup [6,+\infty[} \ (= \mathbb{R} \backslash [0,6])$