Solving x after setting f''(x)=0. help

Hi,

Ok, so f(x)= x^2/((x+1)(x-1))

I am using the second derivative to find the stationary points. But I am really struggling to do so.

The solution says:

"By setting f'(x) = 0 and solving for x, giving x = 0 and x = - 4. The corresponding points on the graph are the relative maximum (0, 0) and the relative minimum at (- 4, 8/9)"

Can someone show me how we get x=0 and x=4, with all the steps.

Thanks.

Mwah

Yes, it's clear that the function f(x) you provide will not give the answers you expect. In these circumstances, I often find that the reason for the discrepancy is that there is a minor typo somewhere in the question. In this case I reverse engineered f(x) to meet the solution provided, and found that this is fully met by taking

f(x) = x²/(x + 1).(x - 2) = x²/(x² - x - 2)

for which df/dx = 2.x/(x² - x - 2) + x².(-1).(2.x - 1)/(x² - x - 2)²

. . . . .= [2.x.(x² - x - 2) - x².(2.x - 1)]/(x² - x - 2)²

. . . . .= -x.(x + 4)/(x² - x - 2)²

which has two stationary points with df/dx = 0 at x = -4 and 0, as required. Substituting these values into f(x) shows that their coordinates are (-4, 8/9) and (0, 0).

For the second differential

d²f/dx² = -(2.x + 4)/(x² - x - 2)² - (x² +4.x).(- 2).(2.x - 1)/(x² - x - 2)³

. . . . . = [-(2.x + 4).(x² - x - 2) + 2.(x² + 4.x).(2.x - 1)]/(x² - x - 2)³

. . . . . = (2.x³ +12.x² + 8)/(x² - x - 2)³

which is positive for x = -4, so the stationary point there is a (very shallow) minimum. At x = 0, d²f/dx² is negative, and that stationary point is a maximum.

HTH

Actually, you use first derivative (not second) to find stationary points.

f(x) = x²/((x+1)(x−1)
...... = x²/(x²−1)

f'(x) = (2x (x²−1) − x² (2x)) / (x²−1)²
...... = (2x³ − 2x − 2x³) / (x²−1)²
...... = −2x / (x²−1)²

Set f'(x) = 0