question 9 (2 points)\nthe function ( f(x)=x^{3}-48 x ) has a local maximum at ( x= ), and has a local…

question 9 (2 points)\nthe function ( f(x)=x^{3}-48 x ) has a local maximum at ( x= ), and has a local minimum at ( x= ). note that here we are looking for the ( x ) value (not the range value). enter a number for each answer.\nview hint for question 9\nquestion 10 (1 point)\nsuppose the second derivative of a function ( f(x) ) is ( f^{prime prime}(x)=2 x-1 ). suppose ( x=1 ) is a critical point of ( f ). using the second derivative test, we know ( x=1 ) is\n- a local maximum\n- a local minimum\n- an absolute minimum\n- an absolute maximum\nview hint for question 10

question 9 (2 points)\nthe function ( f(x)=x^{3}-48 x ) has a local maximum at ( x= ), and has a local minimum at ( x= ). note that here we are looking for the ( x ) value (not the range value). enter a number for each answer.\nview hint for question 9\nquestion 10 (1 point)\nsuppose the second derivative of a function ( f(x) ) is ( f^{prime prime}(x)=2 x-1 ). suppose ( x=1 ) is a critical point of ( f ). using the second derivative test, we know ( x=1 ) is\n- a local maximum\n- a local minimum\n- an absolute minimum\n- an absolute maximum\nview hint for question 10

Answer

Explanation:

Step1: Find the first derivative of (f(x))

The function is (f(x)=x^{3}-48x). Using the power rule ((x^n)^\prime = nx^{n - 1}), the first derivative (f^\prime(x)=3x^{2}-48).

Step2: Find the critical points

Set (f^\prime(x) = 0), so (3x^{2}-48=0). Factor out 3: (3(x^{2}-16)=0), which gives (x^{2}-16=(x + 4)(x - 4)=0). The critical points are (x=-4) and (x = 4).

Step3: Find the second derivative of (f(x))

Differentiate (f^\prime(x)=3x^{2}-48) with respect to (x). Using the power rule, (f^{\prime\prime}(x)=6x).

Step4: Use the second - derivative test

  • For (x=-4): Substitute (x=-4) into (f^{\prime\prime}(x)), (f^{\prime\prime}(-4)=6\times(-4)=-24<0). So (x = - 4) is a local maximum.
  • For (x = 4): Substitute (x = 4) into (f^{\prime\prime}(x)), (f^{\prime\prime}(4)=6\times4 = 24>0). So (x = 4) is a local minimum.

For Question 10:

Step1: Apply the second - derivative test

Given (f^{\prime\prime}(x)=2x - 1) and (x = 1) is a critical point. Substitute (x = 1) into (f^{\prime\prime}(x)), (f^{\prime\prime}(1)=2\times1-1=1>0).

Answer:

Question 9: local maximum at (x=-4), local minimum at (x = 4). Question 10: a local minimum.