(e^x - 1 )/(x^2 - 1) < 0
Any way to get started? I'd appreciate any help.
And also,
xlnx + 1 > x + lnx
(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.
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
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
To solve the inequality (e^x - 1)/(x^2 - 1) < 0, we can start by finding the critical points, which are the points where the numerator and denominator equal zero.
1. Set the numerator equal to zero:
e^x - 1 = 0
To solve for x, add 1 to both sides:
e^x = 1
Take the natural logarithm of both sides:
ln(e^x) = ln(1)
x = 0
So x = 0 is a critical point.
2. Set the denominator equal to zero:
x^2 - 1 = 0
Factor the expression:
(x - 1)(x + 1) = 0
This gives two critical points: x = -1 and x = 1.
Now we can examine the three intervals: (-∞, -1), (-1, 0), and (0, ∞).
1. For x < -1:
Choose a test point, such as x = -2, and substitute it into the inequality:
(e^-2 - 1)/((-2)^2 - 1) < 0
(-1 - 1)/(4 - 1) < 0
-2/3 < 0
Since -2/3 is negative, this interval satisfies the inequality.
2. For -1 < x < 0:
Choose a test point, such as x = -1/2, and substitute it into the inequality:
(e^-1/2 - 1)/((-1/2)^2 - 1) < 0
(sqrt(e) - 1)/(1/4 - 1) < 0
Since sqrt(e) - 1 is positive and 1/4 - 1 is negative, this interval does not satisfy the inequality.
3. For x > 0:
Choose a test point, such as x = 1/2, and substitute it into the inequality:
(e^1/2 - 1)/((1/2)^2 - 1) < 0
(1 - 1)/(1/4 - 1) < 0
Since 1 - 1 is zero and 1/4 - 1 is negative, this interval does not satisfy the inequality.
Therefore, the solution to the inequality (e^x - 1)/(x^2 - 1) < 0 is x < -1.
Moving on to the second problem, xlnx + 1 > x + lnx:
1. Subtract lnx from both sides of the inequality:
xlnx - lnx + 1 - x > 0
2. Factor out lnx and -1 from the left side:
lnx(x - 1) - (x - 1) > 0
(x - 1)(lnx - 1) > 0
Now we consider two cases:
Case 1: (x - 1) > 0
For x > 1, both factors (x - 1) and (lnx - 1) are positive. Therefore, this interval satisfies the inequality.
Case 2: (x - 1) < 0
For x < 1, both factors (x - 1) and (lnx - 1) are negative. Therefore, this interval does not satisfy the inequality.
Therefore, the solution to the inequality xlnx + 1 > x + lnx is x > 1.