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secondary 4 | A Maths
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I’m not sure how to do part (ii). I’ve not checked if my (i) is correct
Date Posted: 4 years ago
Views: 217
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done
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clear
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Girl, the difficulty here lies in understand the meaning of “minimum gradient” when usually questions ask for “minimum point”. Let me know if you need more explanation.
Date Posted:
4 years ago
I got it ! Im just not accustomed to doing d3y/dx3 in questions like this, snd usually stop at d2y/dx2. Ive never done a question like this before so thanks for helping out :D
Girl, can think of the gradient equation of the curve (dy/dx = ...) as a separate variable z, where z = dy/dx.
z = x^2 - 16/x^2
From here, we find the minimum value of z, which is dz/dx = 0.
The second derivative test here is d2z/dx2, which corresponds to the original d3y/dx3.
z = x^2 - 16/x^2
From here, we find the minimum value of z, which is dz/dx = 0.
The second derivative test here is d2z/dx2, which corresponds to the original d3y/dx3.
I’ll think about that . Thank you ☺️
Thank you :)
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