Let f(x) be continuous on [0,6] and differentiable on (0,6). Let f(0)=12 and f(6)=-4 If g(x)=(f(x)/x+1), then for some Lagrange's constant c ∈(0,6), g prime(c)=

Question

Let f(x) be continuous on [0,6] and differentiable on (0,6). Let f(0)=12 and f(6)=-4 If g(x)=(f(x)/x+1), then for some Lagrange's constant c ∈(0,6), g prime(c)= (A) -(44/3) (B) -(22/21) (C) (32/21) (D) -(44/21)

Tardigrade Admin · Accepted Answer

According to Lagrange's Mean value theorem g prime(c) =(g(6)-g(0)/6-0)=((f(6)/6+1)-(f(0)/0+1)/6) =(-(4/7)-12/6)=-(88/42)=-(44/21)

Q. Let $f (x)$ be continuous on $[0, 6]$ and differentiable on $(0, 6)$ . Let $f (0) = 12$ and $f (6) = - 4$ If $g (x) = \frac{f ( x )}{x + 1}$ , then for some Lagrange's constant $c \in (0, 6), g^{'} (c) =$

$- \frac{44}{3}$

$- \frac{22}{21}$

$\frac{32}{21}$

$- \frac{44}{21}$

According to Lagrange's Mean value theorem
$g^{'} (c) = \frac{g ( 6 ) - g ( 0 )}{6 - 0} = \frac{\frac{f ( 6 )}{6 + 1} - \frac{f ( 0 )}{0 + 1}}{6}$
$= \frac{- \frac{4}{7} - 12}{6} = - \frac{88}{42} = - \frac{44}{21}$

Q. Let f(x) be continuous on [0,6] and differentiable on (0,6). Let f(0)=12 and f(6)=−4 If g(x)=x+1f(x)​, then for some Lagrange's constant c∈(0,6),g′(c)=

−344​

−2122​

2132​

−2144​