1. ## Differentiation question

Question: Show that, when k is constant, the curve

$\displaystyle y = 3x^4- 8x^3-6x^2+24x+k$
has a stationary point when x =1 and find the values of x at the other two stationary points on the curve. Find the values of k for which the curve touches the x axis.

My attempt:
$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=12x^{3}-24x^{2}-12x+24$
when $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=12x^{3}-24x^{2}-12x+24 = 0$

$\displaystyle 12x^{3}-24x^{2}-12x+24 = 0$

$\displaystyle 12(x^{3}-2x^{2}-x+2) = 0$
try x = 1

$\displaystyle 12(1^{3}-2(1)^{2}-1(1)+2) = 0$ Thus since f'(1) =0 thus at x =1 there is a stationary point.

The other values of x are x =-1 and x =2

my problem is how to find the values of k

thank you

2. ## Re: Differentiation question

Firstly, think about what the curve looks like.

1. It is a quartic with three stationary points at x = -1, 1 and 2. What sort of stat pts are they, ie max, min, phi?

2. What happens to the curve as x->infinity and x-> - infinity ie on your graph, does it "start" in the top left corner or bottom left corner? Does it end up in the top right corner or the bottom right corner?

That should give you the general shape of the graph.

3. What does "touches" the x-axis mean? At what points could your graph touch the x-axis?

Can you see what to consider next?

3. ## Re: Differentiation question

Nevermind...wasn't paying attention.