Minima maxima: a²y = x⁴ Member for 6 years Submitted by Francis June E… on Wed, 04/08/2020 - 14:34 How to solve a²y = x⁴ Tags Maxima minima Log in or register to post comments Your question don't have Member for 17 years 6 months 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 or register to post comments Yung Ans. Niya po is (0,0) at Member for 6 years 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 or register to post comments Yung process po sana kung Member for 6 years 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 or register to post comments $a^2 y = x^4$ Member for 17 years 6 months 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 or register to post comments Salamat po sir Member for 6 years 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 or register to post comments Please create another forum Member for 17 years 6 months 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 or register to post comments Find the maxima and minima Member for 2 years 2 months Aishann Wed, 01/24/2024 - 21:39 Find the maxima and minima point of the curve y=3x⁴-8x³+6x² Log in or register to post comments
Your question don't have Member for 17 years 6 months 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 or register to post comments
Yung Ans. Niya po is (0,0) at Member for 6 years 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 or register to post comments
Yung process po sana kung Member for 6 years 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 or register to post comments
$a^2 y = x^4$ Member for 17 years 6 months 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 or register to post comments
Salamat po sir Member for 6 years 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 or register to post comments
Please create another forum Member for 17 years 6 months 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 or register to post comments
Find the maxima and minima Member for 2 years 2 months Aishann Wed, 01/24/2024 - 21:39 Find the maxima and minima point of the curve y=3x⁴-8x³+6x² Log in or register to post comments
Your question don't have
Member for
17 years 6 monthsYour 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
Member for
6 yearsIn reply to Your question don't have by Jhun Vert
Yung Ans. Niya po is (0,0) at Minimum.
Yung process sana po sir. Thanks po
Yung process po sana kung
Member for
6 yearsIn reply to Your question don't have by Jhun Vert
Yung process po sana kung paano solve.
Yung ans nya po is (0,0) , minimum.
$a^2 y = x^4$
Member for
17 years 6 months$a^2 y = x^4$
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
Member for
6 yearsIn reply to $a^2 y = x^4$ by Jhun Vert
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
Member for
17 years 6 monthsIn 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
Member for
2 years 2 monthsFind the maxima and minima point of the curve y=3x⁴-8x³+6x²