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Question
The least value of the function f(x) = `"a"x + "b"/x` (where a > 0, b > 0, x > 0) is ______.
Solution
The least value of the function f(x) = `"a"x + "b"/x` (where a > 0, b > 0, x > 0) is `2sqrt("ab")`.
Explanation:
Here, f(x) = `"a"x + "b"/x`
⇒ f'(x) = `"a" - "b"/x^2`
For maximum and minimum value f'(x) = 0
∴ `"a" - "b"/x^2` = 0
⇒ `x^2 = "b"/"a"`
⇒ x = `+- sqrt("b"/"a")`
Now f"(x) = `(2"b")/x^3`
`"f''"(x)_(x = sqrt("b"/"a")) = (2"b")/(("b"/"a")^(3/2))`
= `2 ("a"^(3/2))/("b"^(1/2)) > 0`
Hence, minima
So the least value of the function at x = `sqrt("b"/"a")` is
`"f"(sqrt("b"/"a")) = "a" * sqrt("b"/"a") + "b"/sqrt("b"/"a")`
= `sqrt("ab") + sqrt("ab")`
= `2sqrt("ab")`
Hence, least value is `2sqrt("ab")`.
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