TOUGH c3 exam QU

(a) Show that, for all non-zero values of the constant k, the curve

y= (kx^2 - 1)/(kx^2 + 1)

has exactly one stationary point.

(b) Show that, for all non-zero values of the constant m, the curve

y= e^mx(x^2 + mx)

has exactly two stationary points.

What have you done so far then? Tried differentiation yet?

I havent done anything. I don't understand where to start I dont like the way the question is worded

(a)y=1 - 2/(kx^2+1)

Only one stationary point at x=0

(a) Show that, for all non-zero values of the constant k, the curve

y= (kx^2 - 1)/(kx^2 + 1)

has exactly one stationary point.

(b) Show that, for all non-zero values of the constant m, the curve

y= e^mx(x^2 + mx)

has exactly two stationary points.

(a) Show that, for all non-zero values of the constant k, the curve

y= (kx^2 - 1)/(kx^2 + 1)

has exactly one stationary point.

(b) Show that, for all non-zero values of the constant m, the curve

y= e^mx(x^2 + mx)

has exactly two stationary points.

Could anyone help me with this question?

I've tried to do the quotient rule. I equated it to 0, but I'm not getting an answer, I'm just left with -1 = 1?

Show your working! It should take about one line of algebra once you realize that you only need to worry about the numerator of the derivative

that's what I thought.

Let's see. I got this using quotient rule

$[br]\dfrac{ 2kx(kx^2+1)-2kx(kx^2-1) }{doesn't matter} = 0[br][br]2kx(kx^2+1)-2kx(kx^2-1) = 0[br][br]2kx(kx^2+1) = 2kx(kx^2-1)[br][br]1 = -1[br]$

?

How on earth do you get 1 = -1 from that?

You have 2kx(kx^2 + 1) - 2kx(kx^2 - 1) = 0
so 2k^2x^3 + 2kx - 2k^2x^3 + 2kx = 0
so 4kx = 0
and therefore ...

oh... so I should have multiplied out the brackets. Thanks

Not necessarily - you could have factorized and simplified.

I think what happened is that you divided by zero without realising it!!

pretty sure I didn't do that.

I'm not sure what I did wrong first time, I just followed your way instead... If you look at my working out, I can't actually see anything wrong with it. I did skip a step to get to -1 = 1.

Exactly! Just remind us of the step you skipped

\dfrac{ 2kx(kx^2+1)-2kx(kx^2-1) }{doesn't matter} = 0

2kx(kx^2+1) -2kx(kx^2-1) = 0

2kx(kx^2+1) = 2kx(kx^2-1)

kx^2+1 = kx^2-1

1 = -1

As I thought - you've divided by zero (more accurately, you divided by 2kx but you didn't know whether x was zero or not, so you've "lost" the solution you were meant to find!).