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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.

## Find the maxima and minima

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