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

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