How to solve a²y = x⁴

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 Minimum. Yung process sana po sir. Thanks po

Yung process po sana kung paano solve.

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

$a^2 y = x^4$

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

How about 9a²y=x(4a+x)³

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

Thanks in advance sir

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

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

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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 Minimum.

Yung process sana po sir. Thanks po

Yung process po sana kung paano solve.

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

$a^2 y = x^4$

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

How about 9a²y=x(4a+x)³

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

Thanks in advance sir

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

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

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