Question

For the function f(x) = –6x5 + 160x3, determine the interval(s) for which f(x) decreases.

A) (-4,4)
B) (-4,0) U (4, infinity)
C) (-infinity, -4) U (0,4)
D) (-infinity, -4) U (4, infinity)

Answer 1

@texaschic101 can u plz help :)

Answer 2

I am sorry, but I am not good at this type of problem

Answer 3

@ganeshie8 ..can you help

Answer 4

He's offline (: but thanks for your time!

Answer 5

@thomaster

Answer 6

i can help a little

first take derivative, factor out what you can

you should end up with
x^2-16

factor this
and set your terms to equal zero

Answer 7

Differentiate the function f(x) = –6x^5 + 160x^3
dy/dx=-30x^4+480x^2
Take both 30 and x^2 as a common factor and set the derivative to zero.
30x^2(-x^2+16)=0
30x^2=0 or -x^2+16=0

Answer 8

graphing it

Answer 9

\[f(x)=-6x^5+160 x^3\]
\[f \prime(x)=-30x^4+480 x^2\]
f(x) decreases if \[f \prime(x)<0\]
\[-30 x^2(x^2-16)<0\]

now \(x^2>0\)
hence \((x^2-16)>0\)
\(x^2>16\)
\[\left| x \right|>4\]
x<-4 and x>4
\[(-\infty,-4)\cup (4,\infty)\]

Answer 10

then state your intervals

(inf,-4)
(-4,0)
(0,4)
(4,inf)

then pick a number from each of those intervals

plug back into f'x (your derivative eqaution
and see if it is positive or negative in that interval

positive gives increasing
negative decreasing

then you should see your answer

Answer 11

additional
when you take the 2nd derivative of the function and substitute the critical x-values if f" is negative the function is increasing when f" is positive the function is decreasing

Answer 12

thank u all soo much idk who to choose as best answer u were all amazing!!!!!!!!!!!! i think the answer is: (as surjithayer said) (-infinity, -4) U (4,infinity)