Q. 2

# Let f, g and h be functions from R to R. Show that(f + g)oh = foh + goh(f . g)oh = (foh) . (goh)

(i) (f + g)oh = foh + goh

Let us consider ((f + g)oh)(x) = (f + g)(h(x))

= f(h(x)) + g(h(x))

= (foh)(x) + (goh)(x)

= {(foh) + (goh)}(x)

Then, ((f + g)oh)(x) = {(foh) +(goh)}(x) x ϵ R

Therefore, (f + g)oh = foh + goh.

(ii) (f.g)oh = (foh).(goh)

Let us consider ((f.g)oh)(x) = (f.g)(h(x))

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

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

= (fog)(x).(goh)(x)

= {(fog).(goh)}(x)

Then, ((f.g)oh)(x) = {(fog).(goh)}(x) x ϵ R

Therefore, (f.g)oh = (fog).(goh)

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Functions - 0152 mins  Different kind of mappings58 mins  Functions - 0648 mins  Functions - 1156 mins  Some standard real functions61 mins  Quick Revision of Types of Relations59 mins  Range of Functions58 mins  Functions - 0947 mins  Quick Recap lecture of important graphs & functions58 mins  Range of Quadratic/quadratic & linear/Linear functions45 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 