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

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Integral Calculus> Define a function h : [0, 1] ? [-1, 1] as...

Define a function h : [0, 1] ? [-1, 1] as h(x) = sin p( 2x 2-1 ). Using Inverse function Theorem, find\nv\n-1/ 2.\nHint: Show that h(x) is a bijective function.\nd -1\ndx h\nat x = 0 , x =

raju , 10 Years ago

Grade 12

1 Answers

bharat bajaj

Last Activity: 10 Years ago

goes from [-1,1]. $sin p ({2x^{2}-1})$ As x belongs to [0,1], $sin p ({2x^{2}-1})$ = $sin p ({2y^{2}-1})$ Take f(x) = f(y) for some x,y in domain, At x=0, Answer = 0 $\frac{d}{dx} {h^{-1}(x)} = \frac{1}{\frac{d}{dx} h(x))} = \frac{p(4x)}{cos p(2x^{2}-1})}$ THerefore, x = y. Also, for all b in range, there is some a in the domain. Hence, h(x) is a bijective function. $p ({2x^{2}-1})$ = $p ({2y^{2}-1})$ It implies that