Minima maxima: a²y = x⁴
How to solve a²y = x⁴
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Your question don't have
Your question don't have enough details. What are you trying to solve from this equation? Tagging it with Maxima and Minima is not enough information.
Yung Ans. Niya po is (0,0) at
In reply to Your question don't have by Jhun Vert
Yung process po sana kung
In reply to Your question don't have by Jhun Vert
$a^2 y = x^4$
$a^2 y = x^4$
Differentiate
$a^2 y' = 4x^3$
Equate y' = 0 to determine the critical points (maxima or minima)
$a^2 (0) = 4x^3$
$x = 0$
For x = 0
$a^2 y = 0^4$
$y = 0$
Hence,
critical point = (0, 0)
Check the neighboring points to determine whether (0, 0) is minimum or maximum. Set x = ±1:
$a^2 y = (\pm 1)^4$
$y = +\dfrac{1}{a^2}$ ← above (0, 0)
Hence, the point (0, 0) is minimum.
Salamat po sir
In reply to $a^2 y = x^4$ by Jhun Vert
Please create another forum
In reply to Salamat po sir by Francis June E…
Please create another forum post for your 2nd question as moderators won't allow multiple questions in one thread.
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