How to verify the inequality: {(x+1)*lnx}/2>x-1
- print Print
- list Cite
Expert Answers
calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We have to prove that [(x+1)*lnx]/2 > x-1
[(x+1)*lnx]/2 > x-1
=> [(x+1)*lnx] > 2(x-1)
=> ln x > 2(x-1)/(x+1)
=> ln x - 2(x-1)/(x+1) > 0
If we find the derivative of ln x - 2(x-1)/(x+1)
=> 1/x - 2[(x +1)^-1 - (x - 1)*(x +1)^-2]
=> 1/x - [2/(x+1) - 2*(x-1)/(x+1)^2]
=> 1/x - [2(x+1)/(x+1)^2 - 2*(x-1)/(x+1)^2]
=> 1/x - [(2x+2-2x + 2)/(x+1)^2]
=> 1/x - [4/(x+1)^2]
=> [(x +1)^2 - 4x]/(x+1)^2
=> (x^2 +1 +2x - 4x)/(x+1)^2
=> (x-1)^2/x*(x+1)^2
We see that (x-1)^2 is always positive, also x*(x+1)^2 is always positive for positive values of x for which ln x exists.
So (x-1)^2/x*(x+1)^2 > 0
Therefore we have proved that [(x+1)*lnx]/2 > x-1
Related Questions
- Verify if limit of ln(1+x)/x is 1, x-->0
- 1 Educator Answer
- Evaluate the limit of the fraction (f(x)-f(1))/(x-1), if f(x)=1+2x^5/x^2? x->1
- 1 Educator Answer
- Verify the relation f'(x)/f(x)=1+lnx, if f(x)=e^(ln((x)^x))?
- 1 Educator Answer
- Verify if the value of the limit of the function f(x)=(x^2-6x+5)/(x-1) is 4 x-->1
- 1 Educator Answer
- Prove that limit of the function (a^x-1)/x=lna,x->0,using two methods.
- 1 Educator Answer
First, we'll try to re-write the expression in an easier manner:
(x+1)*lnx > 2(x-1)
Now, we'll divide both sides by (x+1):
ln x > 2(x-1)/ (x+1)
We'll subtract the ratio 2(x-1)/ (x+1) both sides:
ln x - 2(x-1)/ (x+1) > 0
We'll note the given expression:
f(x) = lnx - 2(x-1)/(x+1)
We'll have to demonstrate that f(x)>0.
To prove that a function is increasing, we'll have to prove that the first derivative of the function is positive.
We'll calculate the first derivative:
f'(x) = [lnx-2(x-1)/(x+1)]'
The ratio from the expression of the function will be differentiated using the quotient rule.
f'(x) = (1/x)-{[2(x-1)'*(x+1)-2(x-1)*(x+1)']/(x+1)^2}
f'(x) = (1/x)-(2x+2-2x+2)/(x+1)^2
f'(x) = (1/x)-(4)/(x+1)^2
f'(x) = [(x+1)^2-4x]/x*(x+1)^2
f'(x) = (x^2+2x+1-4x)/x*(x+1)^2
f'(x) = (x^2-2x+1)/x*(x+1)^2
f'(x) = (x-1)^2/x*(x+1)^2
We notice that for any real value of x, (x-1)^2>0 and x*(x+1)^2>0, so f(x)>0.
f(x) = lnx-2(x-1)/(x+1)
lnx-2(x-1)/(x+1) > 0 q.e.d
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Student Answers