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\n( f(x)=2 x^{4}+16 x^{3}+33 )\n\na. the function is increasing on ( (-infty, 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.\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the function is decreasing on ( (-infty,-6) ).\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.\n\nselect the correct choice below and, if necessary, fill in the answer boxes to complete your choice.\n(type integers or simplified fractions.)\na. the function has a local minimum ( f(quad)=quad ), and no local maximum.\nb. the function has a local maximum ( f(quad)=quad ) and a local minimum ( f(quad)=quad ).\nc. the function has a local maximum ( f(quad)=quad ), and no local minimum.\nd. the function has no local extrema.
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: Determine the sign of ( f^\prime(x) ) in different intervals
- For ( x<-6), let ( x=-7). Then ( f^\prime(-7)=8\times(-7)^{2}\times(-7 + 6)=8\times49\times(-1)<0 ).
- For ( -6<x<0), let ( x=-1). Then ( f^\prime(-1)=8\times(-1)^{2}\times(-1 + 6)=8\times1\times5>0 ).
- For ( x>0), let ( x = 1). Then ( f^\prime(1)=8\times1^{2}\times(1 + 6)=8\times1\times7>0 ).
Since ( f^\prime(x)>0) for ( x>-6) (except ( x = 0) where ( f^\prime(x)=0) but the function does not change its increasing - decreasing nature at ( x = 0) because ( f^\prime(x)) has the same sign on both sides of ( x = 0) near ( x = 0)), the function is increasing on ( (-6,\infty)).
Since ( f^\prime(x)<0) for ( x<-6), the function is decreasing on ( (-\infty,-6)).
To find the local extrema:
- We use the first - derivative test. Since the function changes from decreasing (( x<-6)) to increasing (( x>-6)), when ( x=-6), ( f(-6)=2\times(-6)^{4}+16\times(-6)^{3}+33=2\times1296-16\times216 + 33=2592-3456+33=-831). And since the function does not change its increasing - decreasing nature at ( x = 0) (the derivative has the same sign on both sides of ( x = 0) near ( x = 0)), there is no local extremum at ( x = 0).
Answer:
The function is increasing on ( (-6,\infty)), decreasing on ( (-\infty,-6)). The function has a local minimum ( f(-6)=-831) and no local maximum. So for the local - extrema part, the answer is A. The function has a local minimum ( f(-6)=-831), and no local maximum.