Let f: R arrow R defined as f(x)=x5+e(x/2)+ex3 and g(x)=f-1(x), then the value of g prime(2), is

Question

Let f: R arrow R defined as f(x)=x5+e(x/2)+ex3 and g(x)=f-1(x), then the value of g prime(2), is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

fC ( x )= x 5+ e x / 2+ e x 3 f prime(x)=5 x4+(1/2) ex / 2+3 x2 ex3>0 ∴ f ( x ) is one one funciton Clearly given function will be onto function as its range is R. ∵ g(f(x))=x ⇒ g prime(f(x)) ⋅ f prime(x)=1 ⇒ g prime(f(x))=(1/f prime(x)) f(x)=2 text when x=0 ∴ g prime(2)=(1/f prime(0))=(1/((1)2))=2

Q. Let f:R→R defined as f(x)=x5+e2x​+ex3 and g(x)=f−1(x), then the value of g′(2), is

1

2

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4

fC(x)=x5+ex/2+ex3 f′(x)=5x4+21​ex/2+3x2ex3>0 ∴f(x) is one one funciton Clearly given function will be onto function as its range is R. ∵g(f(x))=x⇒g′(f(x))⋅f′(x)=1⇒g′(f(x))=f′(x)1​ f(x)=2 when x=0 ∴g′(2)=f′(0)1​=(21​)1​=2

Q. Let $f : R \to R$ defined as $f (x) = x^{5} + e^{\frac{x}{2}} + e^{x^{3}}$ and $g (x) = f^{- 1} (x)$ , then the value of $g^{'} (2)$ , is