Minima maxima: a²y = x⁴

How to solve a²y = x⁴

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$a^2 y = x^4$
 

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.