For y=x^4 -x^5, write the points for which the second derivative is (+)ve, and points for which it is (-)ve???

Question

The graph of y=x^4-x^5 is here: http://ul.hamzahkhan.com/files/3/Graph.bmp

The answer section in the book has the answer as
"positive for x<0.8, negative for x>0.8"

But isn't the second derivative positive when the graph bends upwards? Then it wouldn't it be positive when 0.8 > x > 0? Since this is the section where the graph bends upwards

Also, wouldn't the second derivative be negative when x<0 and x>0.8, since here it bends downwards?

I'm a little confused here and it would be great if someone can make this a little clearer for me.

Answer 1

y = x^4 - x^5
1st derivative y' = 4x^3 - 5x^4
2nd derivative y'' = 12x^2 - 20x^3

This is positive when 12x^2 > 20x^3
or 3 > 5x
i.e. x < 0.6
It is negative when 12x^2 < 20x^3
or x > 0.6

From the graph you can see that the curve is upward for all x < 0.6, including negative values of x as well as 0 < x < 0.6.

When x > 0.6 you can see the graph curve switches to downward.

Answer 2

I agree with Kemmy that the book's answer is incorrect, but I don't agree with her solution. Iain G has the correct solution. There are turning points at x = 0 and x = 0.8 (first derivative 0 in both cases) but the second derivative is positive for all x < 0.6. In addition, the second derivative is 0 at x = 0.

The second derivative is the rate of change of the gradient (first derivative). When x < 0 the gradient is becoming less negative (ie more positive) so the second derivative is positive for x < 0. The gradient stops becoming more positive at x = 0.6 and starts becoming more negative, so the sign of the second derivative changes at x = 0.6, not x = 0.8, which is the maximum of the curve - the curve's gradient starts becoming less positive (ie more negative) before it reaches the maximum.

Answer 3

Your book's answer is incorrect.

From your graph, the turning points are at 0 and 0.8.
If the gradient is sloping upwards (from left to right), it has a positive gradient and vice versa.

y = x^4 - x^5 => dy/dx = 4x^3 - 5x^4

dy/dx > 0
4x^3 - 5x^4 > 0
5x^4 - 4x^3 < 0
x^3(5x - 4) > 0
0 < x < 4/5
0 < x < 0.8

dy/dx < 0
4x^3 - 5x^4 < 0
5x^4 - 4x^3 > 0
x^3(5x - 4) > 0
x > 4/5 and x < 0
x > 0.8 and x < 0

Answer 4

dy/dx = 4x³ - 5x^4
d²y/dx² = 12x² - 20x³
d²y / dx² = 4x².(3 - 5x)
d²y / dx² > 0 when x < 3/5
d²y /dx² < 0 when x > 3/5

Answer 5

you`re confused,,,,!