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secondary 4 | A Maths
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Date Posted: 5 years ago
Views: 324
Kathy, this one must use first derivative test. The second derivative test fails here. Will write the answers later.
How will know when I use first or second here
This one is my personal experience. Equations of the form y = ax^3 + d will have only one stationary point at x = 0 and at this point d2y/dx2 is 0; this fails, so we must use first.
First always works. Second one most of the time, but not always, as the case here has shown. Usually I will do second derivative test if the expression is easy.
First always works. Second one most of the time, but not always, as the case here has shown. Usually I will do second derivative test if the expression is easy.
I write here for the time being
y = x^3 - 1
dy/dx = 3x2
At stat pt
dy/dx = 0
3x2 = 0
x = 0
y = -1
(0, -1)
First derivative test, test for values of dy/dx around x = 0
x = -0.1, dy/dx is +ve
x = 0, dy/dx is 0
x = 0.1, dy/dx is +ve
Up-zero-up
It’s an inflection.
The graph is what you have learnt
in E Maths for the graph of y = ax^n for n = 3, shifted one unit downwards.
y = x^3 - 1
dy/dx = 3x2
At stat pt
dy/dx = 0
3x2 = 0
x = 0
y = -1
(0, -1)
First derivative test, test for values of dy/dx around x = 0
x = -0.1, dy/dx is +ve
x = 0, dy/dx is 0
x = 0.1, dy/dx is +ve
Up-zero-up
It’s an inflection.
The graph is what you have learnt
in E Maths for the graph of y = ax^n for n = 3, shifted one unit downwards.
Just do a quick check on your second derivative, whether on paper or mentally.
From dy/dx = 3x², it's easy to see that d²y/dx² = 6x (without actually writing) and therefore = 0 since your x = 0. So before actually writing down, you can already switch to the first derivative method.
Alternatively, from dy/dx = 3x², we can already tell without checking any values that the gradient changes from positive before x = 0, to 0 at x = 0 and back to positive again after x = 0 (recall how a graph of y = x² looks like) and therefore it's an inflection point
From dy/dx = 3x², it's easy to see that d²y/dx² = 6x (without actually writing) and therefore = 0 since your x = 0. So before actually writing down, you can already switch to the first derivative method.
Alternatively, from dy/dx = 3x², we can already tell without checking any values that the gradient changes from positive before x = 0, to 0 at x = 0 and back to positive again after x = 0 (recall how a graph of y = x² looks like) and therefore it's an inflection point
Kathy, one more thing. The signage of d2y/dx2 is only useful when dy/dx = 0. If let’s say d2y/dx2 > 0 but dy/dx is not equal to 0, we cannot say that the given point is a minimum point. Because it’s not a turning point.
I think you mean the other way round.
Wrote wrongly the first time, updated.
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Find the stationary point by setting dy/dx=0. Observe that the expression is never negative. This means that at a point slightly to the left of the stationary point, the gradient is positive. At a point slightly to the right of the stationary point, the gradient is also positive. Since the shape is /‾‾/, the stationary point is therefore a point of inflexion. Since the x-coordinate is 0, it is also the y-intercept. To find the x-intercept, substitute y=0 into the equation. Plot these 2 points on the axes and join them suitably to obtain your cubic graph.
Date Posted:
5 years ago
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My version as promised
Date Posted:
5 years ago
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