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Define a function h : [0, 1] ? [-1, 1] as h(x) = sin p[(2x^2-1)/2]. Using Inverse function Theorem, find d/dx (1/h) at x=0, x=-1/(square root of 2) Hint: Show that h(x) is a bijective function. d -1 dx h at x = 0 , x =

Define a function h : [0, 1] ? [-1, 1] as h(x) = sin p[(2x^2-1)/2]. Using Inverse function Theorem, find d/dx (1/h) at x=0, x=-1/(square root of 2)
Hint: Show that h(x) is a bijective function.
d -1
dx h
at x = 0 , x =

Grade:11

1 Answers

bharat bajaj IIT Delhi
askIITians Faculty 122 Points
7 years ago 
As x belongs to [0,1], (2x^2 - 1)/2 goes from [-1,1].
Take f(x) = f(y) for some x,y in domain,
sin pi(2x^2-1/2) =sin p(2y^2-1/2)
It implies that 2x^2 - 1 / 2 = 2y^2 - 1 / 2
THerefore, x = y
Also, for all b in range, there is some a in the domain.
Hence, h(x) is a bijective function.
d/dx h^-1(x) = 1/ (d/dx h(x) ) = pi(2x)/ cos pi(2x^2-1/2)
At x=0,
Answer = 0
At x=-1/v2
Answer = -v2 pi

Thanks
Bharat bajaj
IIT Delhi
askiitians faculty

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