For x>0, what value of x gives the least sum of x and its reciprocal? Show work.
x+(1/x) = y
y&prime = 1 - 1/x2
set that equal to zero.
You didn't specify that you were in calculus though, so I'll provide a second way.
x + (1/x) = y
if x = 1, then y = 2
if x = 1/2 then y = 5/4>2
if x = 2 then y = 5/4>2
if x = 1/3 then y = 10/3>2
if x = 3 then y = 10/3>2
you can see that as x gets bigger and smaller then 1 y gets bigger then 2, so when x = 1 you have a minimum value
To find the value of x that gives the least sum of x and its reciprocal, we need to minimize the expression x + 1/x.
Let's first simplify this expression by finding a common denominator:
x + 1/x = (x^2 + 1)/x
To minimize this expression, we need to find the minimum value of (x^2 + 1)/x. To do this, we can find the derivative and set it equal to zero:
d/dx [(x^2 + 1)/x] = [(2x)(x) - (x^2 + 1)] / x^2
= (2x^2 - x^2 - 1) / x^2
= (x^2 - 1) / x^2
Setting the derivative equal to zero:
(x^2 - 1) / x^2 = 0
Multiply both sides by x^2 to get:
x^2 - 1 = 0
Adding 1 to both sides:
x^2 = 1
Taking the square root of both sides (since x > 0):
x = ±1
Since x > 0, we have:
x = 1
So, the value of x that gives the least sum of x and its reciprocal is x = 1.
To find the value of x that gives the least sum of x and its reciprocal, we need to minimize the expression x + 1/x with x > 0.
To do this, we can take the derivative of the expression with respect to x and set it equal to zero to find the critical points. Let's call the expression f(x):
f(x) = x + 1/x
First, let's find the derivative of f(x):
f'(x) = 1 - 1/x^2
Setting f'(x) equal to zero:
1 - 1/x^2 = 0
Adding 1/x^2 to both sides:
1 = 1/x^2
Now, we can take the reciprocal of both sides:
x^2 = 1
Taking the square root of both sides:
x = ±1
However, since we are considering x > 0, we can discard the negative solution.
Therefore, the critical point is x = 1.
Now, we need to check the endpoints of the interval x > 0, which is x → ∞ as x approaches infinity.
When x → ∞, both x and 1/x → ∞. Therefore, the sum of x and its reciprocal also tends to infinity.
So, the only critical point in the interval x > 0 is x = 1, which gives the least sum of x and its reciprocal.
Therefore, the value of x that gives the least sum is x = 1.