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Question
Show that the value in the conclusion of the mean value theorem for `f(x) = 1/x` on a closed interval of positive numbers [a, b] is `sqrt("ab")`
Solution
f(x) = `1/x, ∈ ["a", "b"]`
f'(x) = `- 1/x^2`
By Mean Value Theorem
f'(c) = `("f"("b") - "f"("a"))/("b" - "a")`
`- 1/"c"^2 = (1/"b" - 1/"a")/("b" - "a")`
`- 1/"c"^2 = ((("a" - "b")/("ab")))/("b" - "a")`
`- 1/"c"^2 = - ((("a" - "b")/("ab")))/("b" - "a") = - 1/"ab"`
c2 = ab
c = `+- sqrt("ab")`
c = `sqrt("ab") ∈ ["a", "b"]`
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