find the intervals on which ( f(x) ) is increasing and the intervals on which ( f(x) ) is decreasing. then…

find the intervals on which ( f(x) ) is increasing and the intervals on which ( f(x) ) is decreasing. then sketch the graph. add horizontal tangent lines.\n( f(x)=x^{4}-12 x^{2} )\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the function is increasing on ( (-3,0),(3, infty) ).\n(type your answer using interval notation. use a comma to separate answers as needed.)\nb. the function is never increasing.\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the function is decreasing on \n(type your answer using interval notation. use a comma to separate answers as needed.)\nb. the function is never decreasing

find the intervals on which ( f(x) ) is increasing and the intervals on which ( f(x) ) is decreasing. then sketch the graph. add horizontal tangent lines.\n( f(x)=x^{4}-12 x^{2} )\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the function is increasing on ( (-3,0),(3, infty) ).\n(type your answer using interval notation. use a comma to separate answers as needed.)\nb. the function is never increasing.\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the function is decreasing on \n(type your answer using interval notation. use a comma to separate answers as needed.)\nb. the function is never decreasing

Answer

Explanation:

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

Use the power rule ( (x^n)^\prime=nx^{n - 1} ). If ( f(x)=x^{4}-12x^{2} ), then ( f^\prime(x)=4x^{3}-24x=4x(x^{2}-6)=4x(x-\sqrt{6})(x + \sqrt{6}))

Step2: Find the critical points

Set ( f^\prime(x)=0 ). ( 4x(x-\sqrt{6})(x+\sqrt{6}) = 0) gives ( x=-\sqrt{6},0,\sqrt{6})

Step3: Test the intervals

  • For the interval ( (-\infty,-\sqrt{6}) ), let ( x=-3). Then ( f^\prime(-3)=4\times(-3)\times((-3)^{2}-6)=4\times(-3)\times3=- 36<0)
  • For the interval ( (-\sqrt{6},0) ), let ( x=-1). Then ( f^\prime(-1)=4\times(-1)\times((-1)^{2}-6)=4\times(-1)\times(-5) = 20>0)
  • For the interval ( (0,\sqrt{6}) ), let ( x = 1). Then ( f^\prime(1)=4\times1\times(1^{2}-6)=4\times1\times(-5)=-20<0)
  • For the interval ( (\sqrt{6},\infty) ), let ( x = 3). Then ( f^\prime(3)=4\times3\times(3^{2}-6)=4\times3\times3 = 36>0)

Answer:

The function ( f(x)=x^{4}-12x^{2}) is increasing on ( (-\sqrt{6},0)\cup(\sqrt{6},\infty)) and decreasing on ( (-\infty,-\sqrt{6})\cup(0,\sqrt{6}))