Q.2 Let f, g and h be functions from R to R. Show that
        \\(f + g) o h = foh + goh\\ (f \cdot g) o h = (foh) \cdot (goh)

Answers (1)

To prove : \\(f + g) o h = foh + goh

                   ((f + g) o h)(x)

                  =(f + g) ( h(x) )

                 =f ( h(x) ) +g(h(x))

                 =(f o h)(x) +(goh)(x) 

               =\left \{ (f o h) +(goh) \right \}(x)                      x\forall R

Hence, \\(f + g) o h = foh + goh

 

 

To prove:(f \cdot g) o h = (foh) \cdot (goh)

                         ((f . g) o h)(x)

                  =(f . g) ( h(x) )

                 =f ( h(x) ) . g(h(x))

                 =(f o h)(x) . (goh)(x) 

              =\left \{ (f o h) .(goh) \right \}(x)                      x\forall R

           \therefore   (f \cdot g) o h = (foh) \cdot (goh)

Hence, (f \cdot g) o h = (foh) \cdot (goh)

 

            

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