Question

Let f(x) be a function such that f '(x)=x(ln|x|)^2. Find the open interval(s) in which is f(x) is concave down in interval notation.

Answer 1

http://mathsfirst.massey.ac.nz/Calculus/Sign2ndDer/images/ConcaveUpDown.png

Answer 2

what do you notice about the tangents to the curve in blue region ?

Answer 3

The tangent lines first went up and then went down

Answer 4

Right, more importantly notice that the "slope" of the tangent is decreasing as x is increased

Answer 5

The tangent keeps falling to the right in the blue region

Answer 6

Since f'(x) gives the slope of the tangent, can we say that f'(x) must be decreasing in the blue region ?

Answer 7

f(x) is concave down
implies f'(x) is decreasing

Answer 8

f'(x) is decreasing
implies f''(x) is less than 0

Answer 9

you're given f'(x)
take the derivative and see in what intervals it is less than 0

Answer 10

f '(x)=x(ln|x|)^2

f''(x) = ?

Answer 11

f''(x)= ln|x|(ln|x|+2) ?

Answer 12

Yes, solve
ln|x|(ln|x|+2) < 0

Answer 13

I'm not sure how to solve that, but here's my answer:
ln|x|<-2

Answer 14

http://www.wolframalpha.com/input/?i=solve++ln |x|%28ln|x|%2B2%29+%3C+0+over+reals

Answer 15

AB < 0

can be split into two cases :

1) A < 0 and B > 0
2) A > 0 and B < 0

Answer 16

How not sure what you mean?

Answer 17

@ganeshie8 how do you find the open intervals with ln|x|<-2??

Answer 18

do you know how to solve?
\[\ln|x| \cdot (\ln|x|+2)=0 \]

Answer 19

Not really..

Answer 20

If A*B=0 then A=0 or B=0 or both=0

Answer 21

do you know how to solve ln|x|=0?

Answer 22

and solve also ln|x|+2=0?

Answer 23

\[\log_a(x)=y \implies a^y=x \\ \text{ assuming } a \in (0,1 ) \cup (1,\infty),x >0\]

Answer 24

\[\ln(x)=\log_e(x)\]

Answer 25

have you solved the above equations I asked about?

Answer 26

I don't quite get it.... e^y=x
Answer: 1, -1??

Answer 27

\[\ln|x|=0 \implies x=1 \text{ or } x=-1 \]

Answer 28

\[\ln|x|=-2 \implies x=? \text{ or } x=?\]

Answer 29

if the absolute value is throwing you
write it like this
assume x>0 then ln|x|=ln(x)
solve ln(x)=0 and also solve ln(x)=-2


then assume x<0 then ln|x|=ln(-x)
solve ln(-x)=0 and solve ln(-x)=-2

Answer 30

\[\ln(x)=0 \implies x=e^{0} \implies x=1 \\ \ln(-x)=0 \implies -x=e^{0} \implies -x=1 \implies x=-1\]

Answer 31

you can solve the other equation in a very similar way

Answer 32

\[\ln|x|=-2 \text{ gives us two equations } \ln(x)=-2 \text{ or } \ln(-x)=-2 \]
can you solve these two equations?

Answer 33

I think it got it:
Answer: 1/e^2 and -1/e^2 ?

Answer 34

looks great

this should be the correct order on the number line of when our f'' is 0