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Let f and g be real functions, defined by `f(x)=(1)/((x+4))andg(x)=(x+4)^(3)`. Find (i) `(f+g)(x)` (ii) `(f-g)(x)` (iii) `(fg)(x)` (iv) `((f)/(g))(x)` (v) `((1)/(f))(x)` |
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Answer» Clearly, `f(x)=(1)/((x+4))` is defined for all real values of x except that at which `x+4=0,i.e.,x=-4`. `:."dom "(f)=R-{-4}`. And, `g(x)=(x+4)^(3)` is defined for all `x""inR`. So, dom (g)=R. `:."dom "(f)nn"dom "(g)=R-{-4}nnR-{-4}`. (i) `(f+g):R-{-4}toR` is given by `(f+g)(x)=f(x)+g(x)=(1)/((x+4))+(x+4)^(4)=(1+(x+4)^(4))/((x+4))` (ii) `(f-g):R-{-4}toR` is given by `(f-g)(x)=f(x)-g(x)=(1)/((x+4))xx(x+4)^(3)=(1-(x+4)^(4))/((x+4))` (iii) `(fg):R-{-4}toR` is given by `(fg)(x)=f(x).g(x)=(1)/((x+4))xx(x+4)^(3)=(x+4)^(2)`. (iv) `{x:g(x)=0}={x:(x+4)^(3)=0}={x:x+4=0}={-4}`. `:."dom "((f)/(g))="dom "(f)nn"dom "(g)-{x:g(x)=0}=R-{-4}`. `((f)/(g))(x )=(f(x))/(g(x))=((1)/((x+4)))/((x+4))=(1)/((x+4)^(4)),xne-4` (v) Clearly, `f(x)ne0" for any "x""inR-{-4}`. `:.(1)/(f):R-{-4}toR` is given by `((1)/(f))(x)=(1)/(f(x))=(1)/((1)/((x+4)))=(x+4)`. |
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