M252: HW 1

HW 1

28a. \[f(x) = 2x^2 \qquad f^{-1}(x) = \sqrt{x/2} \]

28b. The two graphs intersect at \( (0,0) \) and \( (1/2,1/2) \). At \( x = 0 \), \( f(x) \) has slope \( 0 \) and \( f^{-1}(x) \) has slope \( \infty \).

32. \(1/6 \)

42. It is clear \( f(x)\) is everywhere decreasing so it is one-to-one. Its inverse is \( [- (x-1)/8]^{1/3} \).

48. \( g \) must be one-to-one. We can show this by showing the contrapositive statement is true. The contrapositive says that if \( g \) is not one-to-one, then \( f \circ g \) is not one-to-one. To see this suppose \( g \) is not one-to-one. Then there are numbers \( a \ne b \) such that \( g(a) = g(b) \). But then \( f(g(a)) = f(g(b)) \) so \( f \circ g \) is not one-to-one.