A natural log inequality with absolute value

AI Thread Summary
The function F(x) = (8-12ln|x|)/(x^4) is defined for x in the intervals (-∞, 0) and (0, ∞). The expression F(x) equals zero at x = e^(2/3) and x = -e^(2/3). The intervals where F(x) is defined and non-zero are (-∞, -e^(2/3)), (-e^(2/3), 0), (0, e^(2/3)), and (e^(2/3), ∞). By analyzing the sign of F(x) in these intervals, it is determined that F(x) > 0 in the intervals (-e^(2/3), 0) and (0, e^(2/3)). The final solution to the inequality is expressed as (-e^(2/3), 0) U (0, e^(2/3)).
gabrooo
Messages
2
Reaction score
0

Homework Statement



F(x) = (8-12ln|x|)/(x^4) > 0

(a) For what values of x is the expression F(x) defined?
Write your answer in interval notation.

(b) At what value(s) of x is the expression F(x) equal to zero?
If there is more than one answer separate them by commas.

(c) The set of all real numbers x for which the expression
F(x) is defined and non-zero can be written as the union of several
mutually disjoint open intervals. Find this set and express
it as such union.

(d) By analyzing the sign of F(x) on the above open intervals,
solve the inequality expressing your answer in interval notation.

2. The attempt at a solution

now for (a), i guess since we have |x|, so the domain would be (-inf,0)U(0,inf)
for (b), i got ln|x| = 2/3 or |x| = e^(2/3) how to get rid of ||?
(c) .. no idea. and (d) is based on (c).

Please HELP!
 
Physics news on Phys.org
About the point

<br /> \ln |x| = \frac 2 3<br />

The positive number whose logarithm is 2/3 is

<br /> e^{2/3}<br />

As a hint to answering your question: this fact means that the x values you seek solve

<br /> |x| = e^{2/3}<br />


To answer 'c', use the two numbers found above, together with x = 0, to find the three intervals mentioned in part 'c'. Once you know the sign of F at a single spot on each interval, it has that sign throughout, so you can answer the main question.
 
first of all thanks for ure help. i didnt get any examples involving ln|x| inequalities anywhere. if u can tell me whre to see such examples, tht wud be nice
anywys.. here's wht i did:

b)
so, if |x| = e^(2/3)
then x = e^(2/3), -e^(2/3)
these will be values which make F(x) = 0. right?

c)so my intervals for which F(x) is "defined and non-zero" wud be (-inf, -e^(2/3)) U (-e^(2/3), 0) U (0, e^(2/3)) U (e^(2/3), inf)

d) signs wud be - + + -
and i will choose + signs because F(x) > 0. so answer is (-e^(2/3), 0) U (0, e^(2/3))

it wud be nice if u can confirm it! :)
THANKS AGAIN!
 
Looks good.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top