find the x - values of all points where the function has any relative extrema. find the value(s) of any…

find the x - values of all points where the function has any relative extrema. find the value(s) of any relative extrema.\nf(x)=x^{4}-32x^{2}+256\nselect the correct choice below and, if necessary, fill in any answer boxes within your choice.\na. there are no relative maxima. the function has a relative minimum of □ at x = □.\n(use a comma to separate answers as needed.)\nb. there are no relative minima. the function has a relative maximum of □ at x = □.\n(use a comma to separate answers as needed.)\nc. the function has a relative maximum of □ at x = □ and a relative minimum of □ at x = □.\n(use a comma to separate answers as needed.)\nd. there are no relative extrema.

find the x - values of all points where the function has any relative extrema. find the value(s) of any relative extrema.\nf(x)=x^{4}-32x^{2}+256\nselect the correct choice below and, if necessary, fill in any answer boxes within your choice.\na. there are no relative maxima. the function has a relative minimum of □ at x = □.\n(use a comma to separate answers as needed.)\nb. there are no relative minima. the function has a relative maximum of □ at x = □.\n(use a comma to separate answers as needed.)\nc. the function has a relative maximum of □ at x = □ and a relative minimum of □ at x = □.\n(use a comma to separate answers as needed.)\nd. there are no relative extrema.

Answer

Explanation:

Step1: Find the first derivative

Given ( f(x)=x^{4}-32x^{2}+256 ). Using the power rule ( (x^{n})^\prime = nx^{n - 1} ), we have ( f^\prime(x)=4x^{3}-64x=4x(x^{2}-16)=4x(x - 4)(x + 4) ).

Step2: Find the critical points

Set ( f^\prime(x)=0 ), so ( 4x(x - 4)(x + 4)=0 ). Solving for ( x ), we get ( x=-4,0,4 ).

Step3: Find the second derivative

Differentiate ( f^\prime(x)=4x^{3}-64x ) using the power rule. ( f^{\prime\prime}(x)=12x^{2}-64 ).

Step4: Use the second - derivative test

  • For ( x = - 4 ): ( f^{\prime\prime}(-4)=12\times(-4)^{2}-64=192 - 64 = 128>0 ). So ( x=-4 ) is a relative minimum.
  • For ( x = 0 ): ( f^{\prime\prime}(0)=12\times0^{2}-64=-64<0 ). So ( x = 0 ) is a relative maximum.
  • For ( x = 4 ): ( f^{\prime\prime}(4)=12\times4^{2}-64=192 - 64 = 128>0 ). So ( x = 4 ) is a relative minimum.

Answer:

B. There are no relative minima. The function has a relative maximum of ( 256 ) at ( x = 0 )