Find a positive number such that the sum of the number and its reciprocal is as small as possible.
let the number be x
and the sum as described be S
S = x + 1/x
dS/dx = 1 - 1/x^2
= 0 for a max/min of S
1 = 1/x^2
x = ±1 , but x had to be positive
if x=1, then S = 1+1/1 = 2
test: let x = 1.01, S = 1.01 + 1/1.01 > 2
let x= .99, then S = .99 + 1/.99 = 2.000101 > 2
so the number is 1
To find a positive number such that the sum of the number and its reciprocal is as small as possible, we can use calculus.
Let's assume the positive number is x.
The sum of the number and its reciprocal is given by x + 1/x.
To find the minimum value of this sum, we need to find its derivative with respect to x and set it equal to zero.
Let's calculate the derivative:
d/dx(x + 1/x) = 1 - 1/x^2
Now, set the derivative equal to zero and solve for x:
1 - 1/x^2 = 0
Adding 1/x^2 to both sides gives:
1 = 1/x^2
Multiplying both sides by x^2 gives:
x^2 = 1
Taking the square root of both sides gives:
x = ±1
Since we are looking for a positive number, the solution is x = 1.
Therefore, the positive number that minimizes the sum of the number and its reciprocal is 1.
To find a positive number such that the sum of the number and its reciprocal is as small as possible, we can use calculus.
Let's denote the positive number we are looking for as x. We are trying to minimize the sum, which can be expressed as x + 1/x.
To minimize the sum, we need to find the critical points, where the derivative of the sum with respect to x is equal to zero.
Taking the derivative of the sum, we get:
d/dx (x + 1/x) = 1 - 1/x^2
Setting this equal to zero and solving for x:
1 - 1/x^2 = 0
1 = 1/x^2
x^2 = 1
x = ±1
Since we are looking for a positive number, we only consider the positive root:
x = 1
Therefore, the positive number that minimizes the sum of the number and its reciprocal is 1.
If you verify the sum, you will see that 1 + 1/1 = 1 + 1 = 2, which is the smallest possible value for the sum.