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प्रश्न
Using the Lagrange’s mean value theorem determine the values of x at which the tangent is parallel to the secant line at the end points of the given interval:
`f(x) = x^3 - 3x + 2, x ∈ [-2, 2]`
उत्तर
f(x) = x3 – 3x + 2, x ∈ [– 2, 2]
f(x) is continuous in [– 2, 2]
f(x) is differentiable in (– 2, 2)
f(– 2) = (– 2)3 – 3 (– 2) + 2
= – 8 + 6 + 2
= 0
f(2) = (2)3 – 3(2) + 2
= 8 – 6 + 2
= 4
∴ f(x) is defined in the given interval.
Given that tangent is parallel to the secant line of the curve between x = – 2 and x = 2.
∴ f'(c) = `("f"("b") - "f"("a"))/("b" - "a")`
Where f'(x) = 3x2 – 3
3c2 – 3 = `(4 - 0)/(2 + 2)` = 1
3c2 = 4
c2 = `4/3`
c = `+- 2/sqrt(3) ∈ [– 2, 2]`
∴ x = `+- 2/sqrt(3)`
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