The value of c in (0, 2) satisfying the mean value theorem for the function f(x) = x(x - 1)2, x ∈ [0, 2] is equal to

Question

The value of c in (0, 2) satisfying the mean value theorem for the function f(x) = x(x - 1)2, x ∈ [0, 2] is equal to (A) (3/4) (B) (4/3) (C) (1/3) (D) (2/3)

Tardigrade Admin · Accepted Answer

f(x) =x(x-1)2 ; x∈[0,2] f'(c) = (f(b)-f(a)/b-a) ; f(2) = 2, f(1) = 0 f'(x) = 3x2 -4x+1 ⇒ f'(c) = 3c2 - 4c+1 Thus, 3c2 -4c+1 = (f(2) -f(0)/2-0) = (2-0/2-0) = 1 ⇒ c = (4/3)

Q. The value of $c$ in $(0, 2)$ satisfying the mean value theorem for the function $f (x) = x (x - 1)^{2}, x \in [0, 2]$ is equal to

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{1}{3}$

$\frac{2}{3}$

$f (x) = x (x - 1)^{2}; x \in [0, 2]$
$f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}; f (2) = 2, f (1) = 0$
$f^{'} (x) = 3 x^{2} - 4 x + 1 \Rightarrow f^{'} (c) = 3 c^{2} - 4 c + 1$
Thus, $3 c^{2} - 4 c + 1 = \frac{f ( 2 ) - f ( 0 )}{2 - 0}$
$= \frac{2 - 0}{2 - 0} = 1 \Rightarrow c = \frac{4}{3}$

Q. The value of c in (0,2) satisfying the mean value theorem for the function f(x)=x(x−1)2,x∈[0,2] is equal to

43​

34​

31​

32​