+0

# Solving problem with AM-GM

0
2
4
+8

I'm not sure how to approach this problem. Can someone provide a solution with a detailed explanation? All help is greatly appreciated!

Find the minimum value of $$\frac{x^2}{x-1}$$ for x > 1

Oct 9, 2023

#1
+423
+1

To find the minimum value of x2/(x−1) for x>1, we can use the following steps:

Rewrite the expression as a function of u. We can do this by letting u=x−1. Then, x=u+1, so x2=(u+1)2=u2+2u+1. Substituting this into the original expression, we get:

f(u) = \frac{u^2 + 2u + 1}{u}

Find the critical points of f(u). A critical point of a function is a point where the derivative of the function is equal to zero or undefined. To find the critical points of f(u), we need to find the derivative of f(u) and set it equal to zero.

f'(u) = \frac{2u^2 - 2u + 1}{u^2}

Setting f′(u)=0, we get the equation 2u2−2u+1=0. This equation can be factored as (2u−1)2=0, so the only critical point of f(u) is u=21​.

Evaluate f(u) at the critical point and at the endpoints of the interval of interest. Since x>1, we know that u=x−1>0. Therefore, the interval of interest is 0<21​. The endpoints of this interval are u=0 and u=21​.

f(0) = \frac{1}{0} = \infty \\ f\left(\frac{1}{2}\right) = \frac{\frac{1}{4} + 1 + 1}{\frac{1}{2}} = \frac{9}{4}

Compare the values of f(u) at the critical point and the endpoints of the interval. Since f(u) is continuous for 0≤u≤21​, the minimum value of f(u) must occur at either u=0 or u=21​. Since f(0)=∞ and f(21​)=49​, we know that the minimum value of f(u) occurs at u=21​.

Therefore, the minimum value of x2/(x−1) for x>1 is 9/4.

Oct 9, 2023
#2
+33568
+1

Hmm.  Look at the following:

Some checking is in order bingboy!

Oct 10, 2023
#3
+125697
+1

y =  x^2 / (x -1)

Rewrite as

y = x^2 ( x -1)^(-1)          take the derivative, set to 0

y'  = 2x (x -1)^(-1) - x^2 (x-1)^(-2)  = 0

y'  = (x-1)^(-2) [ 2x ( x -1) - x^2]  =  0

y' =  2x^2 - 2x -x^2   = 0

y' =  x^2  - 2x   = 0

x ( x  -2)  = 0

x = 0  or  x  = 2

At x = 1.5, y =  4.5     at  x  =3, y =  4.5

So ....wnen x > 1,  the min occurs at x = 2    { min = 4  }

Oct 11, 2023