(e^x - 1 )/(x^2 - 1) < 0
That is the same as asking: When are the top and bottom of different signs?
If -1<x<1, bottom negative, otherwise+
If x<0, top negative, otherwise+
so
If x is less than -1, the top is - and the bottom is + and your inequality is satisfied.
If x is between 0 and 1, the bottom is - and the top is + and your inequality is satisfied
Outside those two domains, your inequality is not true.
(e^x - 1 )/(x^2 - 1) < 0
Any way to get started? I'd appreciate any help.
And also,
xlnx + 1 > x + lnx
3 answers
xlnx + 1 > x + lnx
I bet e or e^2 or something is worth trying because we want ln x to be nice
try e
e ln e + 1 > e + ln e ???
e * 1 + 1 > e + 1 ?? nope, they are equal. That is a crossover. So try over e
like e^2
e^2 (2 ln e) + 1 > e^2 + 2 ln e ???
2 e^2 + 1 > e^2 + 2 ???
e^2 > 1 Yes, so x>e works
Now try x < e
graph the whole thing for some values of x
I bet e or e^2 or something is worth trying because we want ln x to be nice
try e
e ln e + 1 > e + ln e ???
e * 1 + 1 > e + 1 ?? nope, they are equal. That is a crossover. So try over e
like e^2
e^2 (2 ln e) + 1 > e^2 + 2 ln e ???
2 e^2 + 1 > e^2 + 2 ???
e^2 > 1 Yes, so x>e works
Now try x < e
graph the whole thing for some values of x
Damon did the first one, I'll try the second
rewrite it as
xlnx - lnx > x-1
lnx(x-1) > x-1
divide by x-1, so
lnx > 1 if x >1
lnx < 1 if x<1
lnx = 1 has solution x = e
so "critical values are x =1 and x = e
testing for x between 0 and 1, 1 and e and greater than e
so testing x> e eg. x = 5, it is true
let x = 1.5 does not work
let x= .5 works!
so the solution is 0<x<1 or 1<x<e
rewrite it as
xlnx - lnx > x-1
lnx(x-1) > x-1
divide by x-1, so
lnx > 1 if x >1
lnx < 1 if x<1
lnx = 1 has solution x = e
so "critical values are x =1 and x = e
testing for x between 0 and 1, 1 and e and greater than e
so testing x> e eg. x = 5, it is true
let x = 1.5 does not work
let x= .5 works!
so the solution is 0<x<1 or 1<x<e
1 answer