Minima maxima: a²y = x⁴

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Francis June E....
Minima maxima: a²y = x⁴

How to solve a²y = x⁴

Jhun Vert

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.

Francis June E....

Yung Ans. Niya po is (0,0) at Minimum.
Yung process sana po sir. Thanks po

Francis June E....

Yung process po sana kung paano solve.

Yung ans nya po is (0,0) , minimum.

Jhun Vert

$a^2 y = x^4$

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.

Francis June E....

Salamat po sir

Yung ans nya po is (a,3a) maximum

Jhun Vert

Please create another forum post for your 2nd question as moderators won't allow multiple questions in one thread.

Aishann

Find the maxima and minima point of the curve y=3x⁴-8x³+6x²

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