Question

Verify LMVT for the given functions f (x) = x (2 - x), x ∈ [0, 1]

Answer 1

Given : f ( x ) = x 2 − 2 x + 3 in the interval [4,6]. We know that a polynomial function is continuous everywhere and also differentiable.

Answer 2

Answer:

f'(c) = 1 ∈ [0, 1]

Step-by-step explanation:

f(x) = x(2- x)

$f(x) = 2x - x^2$

x ∈ [0, 1]

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

f'(x) = 2 - 2x

Now,

f(a) = f(0) = 2× 0 - 0 = 0

f(b) = f(1) = 2× 1 - 1 = 1

Hence function satisfy all the conditions of LMVT on f[0, 1]

Now,

$f'(c) = \dfrac {f(b) - f(a)}{b - a}\\ f'(c)= \dfrac {1 - 0}{1 - 0} = 1$

f'(c) = 1 ∈ [0, 1]