Are you allowed to use Calculus for this problem? If so, a function is increasing where its derivative is 0. If not, draw a graph and look at where the graph is increasing...
A spot of light on a computer screen moves in a horizontal line across the screen. At time t seconds, its distance, x mm, from the left-hand edge of the screen is given, for t>/= 0, by
x=t^3-12t^2+kt,
where k is a positive constant. Find the set of values of k for which x is an increasing function of t. (answer: k>/=48)
Im not sure whats the right way to go about solving this problem. Can somebody guide me along? Thank you!
First of all, thanks for the help guys!
So Given that - A differentiable function is increasing where its derivative is positive , i tried putting the (derivative of x) > 0.
so..
3t^3-24t+k>0
3t^2-24t>-k
-k<3t^2-24t
then im stuck again..
Am i going in the right direction?
Sorry for the typo!
To clear up:
I took x=t^3-12t^2+kt and differentiated it to get ---> 3t^2-24t+k (corrected from 3t^3-24t+k which was incorrectly written in my last post)
Anyway, by completing the square method, i got :
(x-4)^2=-k/3 +16
Up till here am i doing the right thing?
Well apart from the fact that it is supposed to be $\displaystyle \begin{align*} ( x - 4 ) ^2 > -\frac{k}{3} + 16 \end{align*}$, that's correct, so now
$\displaystyle \begin{align*} \sqrt{ (x - 4)^2 } &> \sqrt{ 16 - \frac{k}{3} } \\ |x - 4| &> \sqrt{ 16 - \frac{k}{3}} \\ x - 4 < - \sqrt{ 16 - \frac{k}{3} } \textrm{ or } x - 4 &> \sqrt{ 16 - \frac{k}{3}} \\ x < 4 - \sqrt{ 16 - \frac{k}{3} } \textrm{ or } x &> 4 + \sqrt{ 16 - \frac{k}{3}} \end{align*}$