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

find the intervals on which ( f(x) ) is increasing, the intervals on which ( f(x) ) is decreasing, and the local extrema.\n( f(x)=2 x^{4}+16 x^{3}+33 )\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the function is increasing on ( (-6, infty) ).\n(type your answer in interval notation. type integers or simplified fractions. 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 in interval notation. type integers or simplified fractions. use a comma to separate answers as needed)\nb. the function is never decreasing.

find the intervals on which ( f(x) ) is increasing, the intervals on which ( f(x) ) is decreasing, and the local extrema.\n( f(x)=2 x^{4}+16 x^{3}+33 )\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the function is increasing on ( (-6, infty) ).\n(type your answer in interval notation. type integers or simplified fractions. 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 in interval notation. type integers or simplified fractions. use a comma to separate answers as needed)\nb. the function is never decreasing.

Answer

Explanation:

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

Using the power rule ((x^n)^\prime=nx^{n - 1}), for (f(x)=2x^{4}+16x^{3}+33), we have (f^\prime(x)=8x^{3}+48x^{2}=8x^{2}(x + 6))

Step2: Find the critical points

Set (f^\prime(x)=0), so (8x^{2}(x + 6)=0). Solving this equation:

  • (8x^{2}=0) gives (x = 0)
  • (x+6=0) gives (x=-6)

Step3: Test the intervals

We consider the intervals ((-\infty,-6)), ((-6,0)) and ((0,\infty))

  • For (x\in(-\infty,-6)), let (x=-7). Then (f^\prime(-7)=8\times(-7)^{2}\times(-7 + 6)=8\times49\times(-1)<0)
  • For (x\in(-6,0)), let (x=-1). Then (f^\prime(-1)=8\times(-1)^{2}\times(-1 + 6)=8\times1\times5>0)
  • For (x\in(0,\infty)), let (x = 1). Then (f^\prime(1)=8\times1^{2}\times(1 + 6)=8\times1\times7>0)

Answer:

The function is decreasing on ((-\infty,-6))