If g is inverse of f and f( x )= x 2+3 x -3( x 0) then g prime(1) equals-

Question

If g is inverse of f and f( x )= x 2+3 x -3( x >0) then g prime(1) equals- (A) (1/2 g(1)+3) (B) -1 (C) (1/5) (D) (-f prime(1)/(f(1))2)

Tardigrade Admin · Accepted Answer

f -1( x )= g ( x ) ⇒ x = f ( g ( x )) Differentiating both sides, 1=f prime(g(x)) g prime(x) ⇒ g prime(x)=(1/f prime(g(x))) Now f prime( x )=2 x +3 So g prime(x)=(1/2 g(x)+3) ⇒ g prime(1)=(1/2(g(1))+3) textgof( x )= x ⇒ g prime( f ( x )) f prime( x )=1 f(x)=1 at x=1 f prime(1)=5 g prime(1) f prime(1)=1 ⇒ g prime(1)=1 / 5

Q. If $g$ is inverse of $f$ and $f (x) = x^{2} + 3 x - 3 (x > 0)$ then $g^{'} (1)$ equals-

$\frac{1}{2 g ( 1 ) + 3}$

$- 1$

$\frac{1}{5}$

$\frac{- f ^{'} ( 1 )}{( f ( 1 ) ) ^{2}}$

Q. If g is inverse of f and f(x)=x2+3x−3(x>0) then g′(1) equals-

2g(1)+31​

−1

51​

(f(1))2−f′(1)​

f−1(x)=g(x)⇒x=f(g(x)) Differentiating both sides, 1=f′(g(x))g′(x)⇒g′(x)=f′(g(x))1​ Now f′(x)=2x+3 So g′(x)=2g(x)+31​⇒g′(1)=2(g(1))+31​ gof(x)=x⇒g′(f(x))f′(x)=1 f(x)=1 at x=1&f′(1)=5 g′(1)f′(1)=1⇒g′(1)=1/5

Q. If $g$ is inverse of $f$ and $f (x) = x^{2} + 3 x - 3 (x > 0)$ then $g^{'} (1)$ equals-

$\frac{1}{2 g ( 1 ) + 3}$

$- 1$

$\frac{1}{5}$

$\frac{- f ^{'} ( 1 )}{( f ( 1 ) ) ^{2}}$