Show that the maximum value of x + (1/x) is less than its minimum valu

1.	Show that the maximum value of x + (1/x) is less than its minimum value.
Answer» Let y = x + (1/x) ∴ dy/dx = 1 - (1/x²) or, dy/dx = ((x² - 1)/x²) For maximum or minimum value of y, dy/dx = 0 ∴ ((1 - x²)/x²) = 0 = 1 - x² = 0 x = 1,-1 Again, d²y/dx² = 2/x³ At, x = 1, d²y/dx² = 2 > 0 So, y has minimum value at x = 1 and minimum value of y = x + (1/x) = 1 + (1/1) = 2 and at, x = -1, d²y/dx² = 2/(-1)³ = -2 < 0 So, y has maximum value at x = -1 and maximum value of y = -1 + (1/-1) = -2 So, minimum value of given function is greater than maximum value.

Show that the maximum value of x + (1/x) is less than its minimum value.

Answer»

Let y = x + (1/x)

∴ dy/dx = 1 - (1/x²)

or, dy/dx = ((x² - 1)/x²)

For maximum or minimum value of y, dy/dx = 0

∴ ((1 - x²)/x²) = 0

= 1 - x² = 0

x = 1,-1

Again, d²y/dx² = 2/x³

At, x = 1, d²y/dx² = 2 > 0

So, y has minimum value at x = 1

and minimum value of y = x + (1/x) = 1 + (1/1) = 2

and at, x = -1,

d²y/dx² = 2/(-1)³ = -2 < 0

So, y has maximum value at x = -1

and maximum value of y = -1 + (1/-1) = -2

So, minimum value of given function is greater than maximum value.

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Show that the maximum value of x + (1/x) is less than its minimum value.

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