Minima maxima: a²y = x⁴ Submitted by Francis June E… on Wed, 04/08/2020 - 14:34 How to solve a²y = x⁴ Tags Maxima minima Log in to post comments Your question don't have Jhun Vert Wed, 04/08/2020 - 15:27 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. Log in to post comments Yung Ans. Niya po is (0,0) at Francis June E… Thu, 04/09/2020 - 09:10 In reply to Your question don't have by Jhun VertYung Ans. Niya po is (0,0) at Minimum. Yung process sana po sir. Thanks po Log in to post comments Yung process po sana kung Francis June E… Thu, 04/09/2020 - 09:15 In reply to Your question don't have by Jhun VertYung process po sana kung paano solve. Yung ans nya po is (0,0) , minimum. Log in to post comments $a^2 y = x^4$ Jhun Vert Thu, 04/09/2020 - 10:22 $a^2 y = x^4$ 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. Log in to post comments Salamat po sir Francis June E… Thu, 04/09/2020 - 16:27 In reply to $a^2 y = x^4$ by Jhun VertSalamat po sir How about 9a²y=x(4a+x)³ Yung ans nya po is (a,3a) maximum Thanks in advance sir Log in to post comments Please create another forum Jhun Vert Thu, 04/09/2020 - 16:39 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. Log in to post comments Find the maxima and minima Aishann Wed, 01/24/2024 - 21:39 Find the maxima and minima point of the curve y=3x⁴-8x³+6x² Log in to post comments
Your question don't have Jhun Vert Wed, 04/08/2020 - 15:27 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. Log in to post comments
Yung Ans. Niya po is (0,0) at Francis June E… Thu, 04/09/2020 - 09:10 In reply to Your question don't have by Jhun VertYung Ans. Niya po is (0,0) at Minimum. Yung process sana po sir. Thanks po Log in to post comments
Yung process po sana kung Francis June E… Thu, 04/09/2020 - 09:15 In reply to Your question don't have by Jhun VertYung process po sana kung paano solve. Yung ans nya po is (0,0) , minimum. Log in to post comments
$a^2 y = x^4$ Jhun Vert Thu, 04/09/2020 - 10:22 $a^2 y = x^4$ 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. Log in to post comments
Salamat po sir Francis June E… Thu, 04/09/2020 - 16:27 In reply to $a^2 y = x^4$ by Jhun VertSalamat po sir How about 9a²y=x(4a+x)³ Yung ans nya po is (a,3a) maximum Thanks in advance sir Log in to post comments
Please create another forum Jhun Vert Thu, 04/09/2020 - 16:39 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. Log in to post comments
Find the maxima and minima Aishann Wed, 01/24/2024 - 21:39 Find the maxima and minima point of the curve y=3x⁴-8x³+6x² Log in to post comments
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