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Question
f(x) = (x-1)(x-2)(x-3) , x ε[0,4], find if 'c' LMVT can be applied
Solution
f(x) = (x-1)(x-2)(x-3) , x ∈[0,4],
∴ f(x) = x3 -6x2 + 11x -6
As f(x) is a polynomial in x
(1) f(x) is continuous on [0, 4]
(2) f(x) is differentiable on (0, 4)
Thus, all the conditions of LMVT are satisfied.
To verify LMVT we have to find c ∈ (0,4) such that
`"f'" ("c") = ("f"(4 )-"f"(0))/(4-0)` ........(1)
Now `"f"(4) = (4-1)(4-2)(4-3) = 6`
f(0) = (0-1)(0-2)(0-3) = -6 and
f '(x) = 3x2 - 12x +11
∴ f '(c) = 3c2 -12c +11
∴ from (1)
`("f"(4) -"f"(0))/(4-0) = 3"c"^2 - 12"c" +11`
`(6-(-6))/4 = 3"c"^2 -12 "c" +11`
3 = 3c2 -12c +11
3c2 -12c +8 =0
c = `("b"+-sqrt("b"^2 - 4"ac"))/"2a"`
c = `(12±sqrt(144 -96))/6`
c =`(12± sqrt48)/6`
c = `12 ±2/3 sqrt3`
c =` 4 ± 2/sqrt3`
Both the value of c lie between 0 and 4.
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