Consider the functions defined implicitly by the equation y3-3 y+x=0 on various intervals in the real line. If x ∈(-∞,-2) ∪(2, ∞), the equation implicitly defines a unique real valued differentiable function y=f(x). If x ∈(-2,2), the equation implicitly defines a unique real valued differentiable function y=g(x) satisfying g(0)=0 ∫ limits-11 g'(x) d x=

Question

Consider the functions defined implicitly by the equation y3-3 y+x=0 on various intervals in the real line. If x ∈(-∞,-2) ∪(2, ∞), the equation implicitly defines a unique real valued differentiable function y=f(x). If x ∈(-2,2), the equation implicitly defines a unique real valued differentiable function y=g(x) satisfying g(0)=0 ∫ limits-11 g'(x) d x= (A) 2 g(-1) (B) 0 (C) -2 g(1) (D) 2 g(1)

Tardigrade Admin · Accepted Answer

We have y'=(1/3(1-(f(x)2))) which is even Hence ∫ limits-11 g'(x)=g(1)-g(-1)=2 g(1)

$2 g (- 1)$

0

$- 2 g (1)$

$2 g (1)$

We have $y^{'} = \frac{1}{3 ( 1 - ( f ( x ) ^{2} ) )}$ which is even
Hence $- 1 \int 1 g^{'} (x) = g (1) - g (- 1) = 2 g (1)$

2g(−1)

0

−2g(1)

2g(1)

We have y′=3(1−(f(x)2))1​ which is even Hence −1∫1​g′(x)=g(1)−g(−1)=2g(1)

$2 g (- 1)$

$- 2 g (1)$

$2 g (1)$

We have $y^{'} = \frac{1}{3 ( 1 - ( f ( x ) ^{2} ) )}$ which is even
Hence $- 1 \int 1 g^{'} (x) = g (1) - g (- 1) = 2 g (1)$