Q5. f(x) = x2 and g(x) = x + 1 on R. Find fog and gof. For which x are they equal?
fog(x) = (x + 1)2 = x2 + 2x + 1 and gof(x) = x2 + 1. They are equal only when 2x = 0, that is at x = 0, so fog ≠ gof.
Q6. f : {1, 3, 4} → {1, 2, 5} is {(1, 2), (3, 5), (4, 1)} and g : {1, 2, 5} → {1, 3} is {(1, 3), (2, 3), (5, 1)}. Write gof as a set of pairs.
gof(1) = g(2) = 3, gof(3) = g(5) = 1 and gof(4) = g(1) = 3. So gof = {(1, 3), (3, 1), (4, 3)}.
Q7. f(x) = (4x + 3)/(6x − 4) for x ≠ 2/3. Show that fof(x) = x for every x ≠ 2/3. What is the inverse of f?
Over the common denominator 6x − 4:
4f(x) + 3 = 34x/(6x − 4),
6f(x) − 4 = 34/(6x − 4).
Dividing, fof(x) = 34x/34 = x. So f undoes itself: f−1 = f. (f(x) is never 2/3, since 3(4x + 3) = 2(6x − 4) would need 9 = −8, so fof is defined.)
Q8. f(x) = x + 1 and g(x) = 2x on R. Find (gof)−1 directly, and check that it equals f−1og−1.
gof(x) = 2x + 2, so (gof)−1(y) = (y − 2)/2. g−1(y) = y/2 and f−1(y) = y − 1, so f−1(g−1(y)) = y/2 − 1 = (y − 2)/2. They agree.
Q9. Find the inverse of f : R → R, f(x) = x3 + 1, and evaluate f−1(28).
x3 is one-one and onto on R, so f is too. From y = x3 + 1, x = ∛(y − 1), so f−1(y) = ∛(y − 1) and f−1(28) = ∛27 = 3.
Q10. For f(x) = √x and g(x) = x − 9, find the largest set of real x on which gof and fog are defined.
gof(x) = √x − 9 needs only x ≥ 0. fog(x) = √(x − 9) needs x − 9 ≥ 0, that is x ≥ 9. So gof is defined on [0, ∞) and fog on [9, ∞).