find the absolute maximum and minimum values of the function over the indicated interval, and indicate the x…

find the absolute maximum and minimum values of the function over the indicated interval, and indicate the x - values at which they occur.\nf(x)=x² - 6x - 4; 0,4\nfind the first derivative of f.\nf(x)=□\n(simplify your answer.)\nthe absolute maximum value is □ at x = □.\n(use a comma to separate answers as needed.)\nthe absolute minimum value is □ at x = □.\n(use a comma to separate answers as needed.)

find the absolute maximum and minimum values of the function over the indicated interval, and indicate the x - values at which they occur.\nf(x)=x² - 6x - 4; 0,4\nfind the first derivative of f.\nf(x)=□\n(simplify your answer.)\nthe absolute maximum value is □ at x = □.\n(use a comma to separate answers as needed.)\nthe absolute minimum value is □ at x = □.\n(use a comma to separate answers as needed.)

Answer

Explanation:

Step1: Find the first derivative

Using the power rule ( (x^n)^\prime=nx^{n - 1} ), for ( f(x)=x^{2}-6x - 4 ), we have ( f^\prime(x)=(x^{2})^\prime-(6x)^\prime-(4)^\prime ). Since ( (x^{2})^\prime = 2x ), ( (6x)^\prime=6 ), and ( (4)^\prime = 0 ), then ( f^\prime(x)=2x-6 ).

Step2: Find the critical points

Set ( f^\prime(x)=0 ), so ( 2x - 6=0 ). Solving for ( x ): Add ( 6 ) to both sides: ( 2x=6 ). Divide both sides by ( 2 ): ( x = 3 ).

Step3: Evaluate the function at the critical point and endpoints

  • For ( x = 0 ): ( f(0)=0^{2}-6\times0 - 4=-4 ).
  • For ( x = 3 ): ( f(3)=3^{2}-6\times3 - 4=9 - 18 - 4=-13 ).
  • For ( x = 4 ): ( f(4)=4^{2}-6\times4 - 4=16 - 24 - 4=-12 ).

Answer:

The first derivative ( f^\prime(x)=2x - 6 ). The absolute maximum value is (-4) at (x = 0). The absolute minimum value is (-13) at (x = 3).