Question

Find c if LMVT is applicable for f (x) = x (x -1) (x - 2), x ∈ [0, 1/2]

Answer 1

it is your answer mate
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Answer 2

Answer:

3/4

Step-by-step explanation:

f(x) = x(x-1)(x- 2)

$f(x) = x(x^2 - 3x + 2 )\\f(x) = x^3 -3x^2 + 2x$

x ∈ [0, 1/2]

function f(x) be continuous on interval [0, 1/2] So f(x) is differential

f'(x) = 3x^2 - 6x + 2[/tex][/tex]

Now,

f(a) = f(0) = 3×0 - 6×0 + 2×0 = 0

$f(b) = f(\dfrac 12)=(\dfrac 12)^3- 3\times (\dfrac12)^2 + 2 \times \dfrac 12\\f(b) = f(\dfrac 12)=\dfrac 18 - \dfrac 34 + 1\\\\f(b) = f(\dfrac 12)=\dfrac {1 - 6 + 8}8\\\\f(b) = f(\dfrac 12)=\dfrac {3}8$

Hence function satisfy all the conditions of Lagrange's mean value theorem on f[0, 1/2]

Now,

$f'(c) = \dfrac {f(b) - f(a)}{b - a}\\\\f'(c)= \dfrac {3/8 - 0}{1/2 - 0}\\f'(c)= \dfrac 34$ ∈ [0, 1/2]

f'(c) = 3/4 ∈ [0, 1/2]