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Question
Show that the value in the conclusion of the mean value theorem for `f(x) = "A"x^2 + "B"x + "C"` on any interval [a, b] is `("a" + "b")/2`
Solution
f(x) = Ax² + Bx + C, x ∈ [a, b]
f'(x) = 2Ax + B
By Mean Value Theorem,
f'(c) = `("f"("b") - "f"("a"))/("b" - "a")` ......[∵ f'(x) = 2Ax + B]
2Ac + B = `(("AB"^2 + "Bb" + "C") - ("Aa"^2 + "Ba" + "C"))/("b" - "a")`
2Ac + B = `("Ab"^2 + "Bb" + "C" - "Aa"^2 - "BA" - "C")/("b" - "a")`
2Ac + B = `("A"("b"^2 - "a"^2) + "B"("b" - "a"))/("b" - "a")`
2Ac + B = `("A"("b" + "a")("b" - "a") + "b"("b" - "a"))/("b" - "a")`
2Ac + B = `(("b" - "a") ["A"("b" + "a") + "B"])/("b" - "a")`
2Ac + B = A(b + a) + B
2Ac = A(a + b)
c = `("a" + "b")/2 ∈ ["a", "b"]`
∴ x = `("a" + "b")/2`
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