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secondary 4 | A Maths
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I don't understand part ii and iii
Date Posted: 4 years ago
Views: 245
Later I do this by 4 am.
Doing this now. I got too tired when I returned yesterday.
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Tara, there is something not right with the question.
dy/dx = -2xe^(-x^2) + 2 if you have done correctly.
The graph of y = e^(-x^2) + 2x does not seem to have any stationary point at all.
dy/dx = -2xe^(-x^2) + 2 if you have done correctly.
The graph of y = e^(-x^2) + 2x does not seem to have any stationary point at all.
Date Posted:
4 years ago
There is nothing wrong with the question
Because I plotted the graph of both y and dy/dx to check the situation. Apprarently the graph has no stationary points, so I checked the graph of dy/dx and indeed there is no value of x for which dy/dx = 0. All the values of dy/dx are positive so there are no stationary points.
1.
Had the equation been y = e*^(-x^2) + 2x^2 instead,
dy/dx = 2xe^(-x^2) + 4x
And the only stationary point is x = 0, when y = 1.
2.
Had the equation been y = e*^(-x^2) + 2 instead,
dy/dx = 2xe^(-x^2)
And the only stationary point is x = 0, when y = 3.
Had the equation been y = e*^(-x^2) + 2x^2 instead,
dy/dx = 2xe^(-x^2) + 4x
And the only stationary point is x = 0, when y = 1.
2.
Had the equation been y = e*^(-x^2) + 2 instead,
dy/dx = 2xe^(-x^2)
And the only stationary point is x = 0, when y = 3.
For part iii, assuming that the stationary point has been found, “by considering the signage of dy/dx” is another way to ask us to use the first derivative test, since we are checking whether the values of dy/dx are positive or negative.
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