The value of c in the Lagrange's mean value theorem for the function f (x)=x3-4x2+8x+11, when x ϵ [0, 1]

Question

The value of c in the Lagrange's mean value theorem for the function f (x)=x3-4x2+8x+11, when x ϵ [0, 1] (A) (4-√7/3) (B) (2/3) (C) (√7-2/3) (D) (4-√5/3)

Tardigrade Admin · Accepted Answer

f (0)=11 f (1)=16 (f (1)-f (0)/1-0)=3c2-8c+8 ⇒ 3c2 - 8c + 8 = 5 ⇒ 3c2 - 8c + 3 = 0 c ∈ [0, 1] ⇒ c=(4-√7/3)

Q. The value of $c$ in the Lagrange's mean value theorem for the function $f (x) = x^{3} - 4 x^{2} + 8 x + 11,$ when $x ϵ [0, 1]$

$\frac{4 - 7}{3}$

$\frac{2}{3}$

$\frac{7 - 2}{3}$

$\frac{4 - 5}{3}$

$f (0) = 11$
$f (1) = 16$
$\frac{f ( 1 ) - f ( 0 )}{1 - 0} = 3 c^{2} - 8 c + 8$
$\Rightarrow 3 c^{2} - 8 c + 8 = 5$
$\Rightarrow 3 c^{2} - 8 c + 3 = 0 < b r / > c \in [0, 1] \Rightarrow c = \frac{4 - 7}{3}$

Q. The value of c in the Lagrange's mean value theorem for the function f(x)=x3−4x2+8x+11, when xϵ[0,1]

34−7​​

32​

37​−2​

34−5​​