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Let f: R-->R and g:R-->R be two function defined by f(x)=(x^2 -1) and g(x)=(sinx+cosx ). then the function (fog)(x) is invertible on the domain .

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Solution:     (fog)(x)=f[g(x)]=f(sin x+cos x)

                     (fog)(x)=(sin x+cos x)^2-1=sin 2x

                     The function    h(x)=sin 2x is bijective on [-fracpi4,fracpi4]

                     and , therefore ,   (fog)(x)  is invertiable .

                     Leftrightarrow      -fracpi2leq 2xleq fracpi2

                     Leftrightarrow        -fracpi4leq xleq fracpi4.

Posted by

Deependra Verma

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