Question

For each of the given functions, find the intervals on which f(x) is increasing, the intervals on which f(x) is decreasing, and the coordinates of all local extrema.

f(x)= 2x^2 - 16ln x

Answer 1

according to the function, x∈(0,+∞), its derivative should be f'(x)=4x-16/x.
let f'(x)=4x-16/x=0, x=2.
the change of x, f'(x), and f(x) is as following:
x (0,2) 2 (2,+∞)
f'(x) - 0 +
f(x) decrease 8-16ln 2 increase
from above, we can see that the interval on which f(x) is increasing is (2,+∞), and the interval on which f(x) is decreasing is (0,2).

Answer 2

what is the local minimum

Answer 3

the local mininum, naturally, is f(2)=8-16ln 2

Answer 4

i got the point (2,-2.5) but she circled the -2.5 and i dont understand that

Answer 5

it's not -2.5, 8-16ln2≈-3.090 , not -2.5

Answer 6

duh...lol

Answer 7

could you help me on one more like this?

Answer 8

\[f(x)= 4x ^{3} + 3x ^{2} +2\]

Answer 9

i'd love to. well, first we should remember the basic derivatives!?
for example, f(x)=x^a, then f'(x)=ax^(a-1).
f(x)=c (c is a constant), then f'(x)=0.
also the algorithm of derivatives, such as
[f(x)±g(x)]'=f'(x)±g'(x).
so f(x)=4x^3+3x^2+2, f'(x)=12x^2+6x

Answer 10

okay i understand that

Answer 11

great. so just now did you want me to work out the intervals that are increasing or decreasing?

Answer 12

yes and the local extrema please

Answer 13

okay. first, it may be a good habit to do a factoring with the derivative (at least my teacher told me like this) because it will be convenient to calculate. f'(x)=12x^2+6x=x(12x+6).
next, we are sure that the derivative at local extrema should be 0. let f'(x)=x(12x+6)=0. so you can work out the solutions of x here very fast, x1=0, x2=-1/2.

Answer 14

notice here that the domain of definition of the function is R. split it into intervals and points according to the solutions of x, as (-∞,-1/2), -1/2, (-1/2,0), 0 , (0,+∞)

Answer 15

okay so far?

Answer 16

hold on working on it

Answer 17

so it is decreasing at 0,0 and increasing at 0,0

Answer 18

the change of x, f'(x), and f(x) is as following:
x (-∞,-1/2) -1/2 (-1/2,0) 0 (0,+∞)
f'(x) + 0 - 0 +
f(x) ↗ 9/4 ↘ 2 ↗

Answer 19

the +/- of the f'(x) stands for whether f'(x) is positive or negative.
when f'(x)>0, increases; when f'(x)<0, decreases.
so it is directly perceived that the increasing intervals are (-∞,-1/2) , (0,+∞); the decreasing interval is (-1/2,0).
the maximum value is f(-1/2)=9/4 and the minimum value is f(0)=2

Answer 20

oh! now i get you

Answer 21

(^ ^) love to communicate with you.

Answer 22

what do you mean?

Answer 23

what?

Answer 24

love to communicate with you?