A function f is differentiable in the interval 0 ≤ x ≤ 5 such that f(0)=4 and f(5)=-1 . If g(x)=(f(x)/x+1), then there exists some c ∈(0,5) such that g'(c)=

Question

A function f is differentiable in the interval 0 ≤ x ≤ 5 such that f(0)=4 and f(5)=-1 . If g(x)=(f(x)/x+1), then there exists some c ∈(0,5) such that g'(c)= (A) -(1/6) (B) -(5/6) (C) (1/6) (D) (5/6)

Tardigrade Admin · Accepted Answer

Clearly g(x) satisfies condition of LMVT ∴ (g(5)-g(0)/5-0)=g'( c ), c ∈(0,5) ⇒ ((f(5)/6)-(f(0)/1)/5)=g'( c ) ⇒ -(5/6)=g'(c)

Q. A function $f$ is differentiable in the interval $0 \leq x \leq 5$ such that $f (0) = 4$ and $f (5) = - 1.$ If $g (x) = \frac{f ( x )}{x + 1},$ then there exists some $c \in (0, 5)$ such that $g^{'} (c) =$

$- \frac{1}{6}$

$- \frac{5}{6}$

$\frac{1}{6}$

$\frac{5}{6}$

Clearly $g (x)$ satisfies condition of LMVT
$∴ \frac{g ( 5 ) - g ( 0 )}{5 - 0} = g^{'} (c), c \in (0, 5)$
$\Rightarrow \frac{\frac{f ( 5 )}{6} - \frac{f ( 0 )}{1}}{5} = g^{'} (c)$
$\Rightarrow - \frac{5}{6} = g^{'} (c)$

Q. A function f is differentiable in the interval 0≤x≤5 such that f(0)=4 and f(5)=−1. If g(x)=x+1f(x)​, then there exists some c∈(0,5) such that g′(c)=

−61​

−65​

61​

65​