If the function f(x)=-4 e(1-x/2)+1+x+(x2/2)+(x3/3) and g(x)=f-1(x), then the value of g prime((-7/6)) equals

Question

If the function f(x)=-4 e(1-x/2)+1+x+(x2/2)+(x3/3) and g(x)=f-1(x), then the value of g prime((-7/6)) equals (A) (1/5) (B) -(1/5) (C) (6/7) (D) -(6/7)

Tardigrade Admin · Accepted Answer

( dy / dx )=(4/2) e (1- x /2)+1+ x + x 2=2 e ((1- x /2))+ x 2+ x +1 g prime( y )=( dx / dy )=(1/2 e -(( x -1)2)+ x 2+ x +1), when y =-(7/6) then x =1 .( dy / dx )] x =-7 / 6=(1/2+3)=(1/5)

Q. If the function $f (x) = - 4 e^{\frac{1 - x}{2}} + 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{3}$ and $g (x) = f^{- 1} (x)$ , then the value of $g^{'} (\frac{- 7}{6})$ equals

$\frac{1}{5}$

$- \frac{1}{5}$

$\frac{6}{7}$

$- \frac{6}{7}$

$\frac{d y}{d x} = \frac{4}{2} e^{\frac{1 - x}{2}} + 1 + x + x^{2} = 2 e^{(\frac{1 - x}{2})} + x^{2} + x + 1$
$g^{'} (y) = \frac{d x}{d y} = \frac{1}{2 e ^{- (\frac{x - 1}{2})} + x ^{2} + x + 1}$ , when $y = - \frac{7}{6}$ then $x = 1$
$\frac{d y}{d x}]_{x = - 7/6} = \frac{1}{2 + 3} = \frac{1}{5}$

Q. If the function f(x)=−4e21−x​+1+x+2x2​+3x3​ and g(x)=f−1(x), then the value of g′(6−7​) equals

51​

−51​

76​

−76​