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MA1S11 Calculus, Tutorial Sheet 21 13-16 October 2015 1. (3) Determine the symmetry properties (symmetry about x-axis, about y-axis, about the origin or none at all) of the following curves in the xy plane (you do not need to draw the graphs!): y 3 = x7 + sin(x), x2 + y2 = 1, 4 |y| = x−3 + x. Solution: We just need to see whether either a separate or a combined change of sign of x, y alters the equation or not. If not we have a symmetry. For instance y 3 = x7 + sin(x) does change for either x → −x or y → −y but not if both are changed simultaneously, so it is symmetric about the origin, but not about the x2 axis nor the y-axis. The equation x2 + y4 = 1 does not change whatever we do to the signs of x or y, so it has all symmetries (about x-, y-axis and origin). In the last case we get the same equation for y → −y, so we have a symmetry about the x-axis. However, changing x → −x changes the equation as does a simultaneous sign change of x and y, so there is no further symmetry. 2. (3) Consider the functions √ g(x) = 2 x2 − 1. f (x) = 3x + 1, What are their natural domains and their ranges? Work out the compositions f ◦ g and g ◦ f with their respective natural domains. Solution: The domain and range for f are the real numbers. For g the domain are the real numbers for which x2 ≥ 1, so we have to exclude the open interval (−1, 1). Varying x over this domain gives all non-negative numbers under the square root, so the range of g are the non-negative real numbers. We have √ (f ◦ g)(x) = f (g(x)) = 3g(x) + 1 = 6 x2 − 1 + 1, p p √ (g ◦ f )(x) = g(f (x)) = 2 (f (x))2 − 1 = 2 (3x + 1)2 − 1 = 2 9x2 + 6x. Clearly, Df ◦g is the same as the domain of g. To find Dg◦f we need to find out for which inputs x the term under the square root is non-negative, i.e. we need to solve the inequality 9x2 + 6x = 3x(3x + 2) ≥ 0 . A product of 2 factors is non-negative if either both factors are non-negative or both factors are non-positive. Hence, either we have simultaneously 3x ≥ 0 and 1 Stefan Sint, sint@maths.tcd.ie, see also http://www.maths.tcd.ie/~sint/MA1S11.html 1 3x + 2 ≥ 0 or we have simultaneously 3x ≤ 0 and 3x + 2 ≤ 0. In the first case x ≥ 0 is the stronger condition and in the second case x ≤ −2/3 is the stronger condition. Therefore Dg◦f = {x : x ≤ −2/3 or x ≥ 0}. √ 3. (2) We define the function f (x) = x3 − 1. What is its natural domain and its range? Is this function invertible? If so, determine the inverse function with its domain and range. Solution: The function is real if the expression under the root is non-negative, i.e. if x3 − 1 ≥ 0 This simplifies to x ≥ 1, so the natural domain are the real numbers ≥ 1. Varying x over the domain one gets all non-negative real numbers, which defines the range of f . The function is strictly monotonically increasing and thus invertible. A graph of the function is given in figure 1 Solving y = f (x) for x, 10 8 6 4 2 K4 K2 0 2 x Figure 1: question 1: graph of f (x) = y= √ 4 √ x3 − 1 ⇔ y 2 + 1 = x3 ⇔ x = and exchanging x and y we have f −1 (x) = √ 3 x3 − 1. p 3 y2 + 1 x2 + 1 = (x2 + 1)1/3 The domain of the inverse function f −1 is the range of f , so all real numbers x ≥ 0, and the range of f −1 coincides with the domain of f , so all numbers y ≥ 1. 2 5 4 3 y 2 1 0 0 1 2 3 x Figure 2: question 1: graphs of f (x) = √ 4 5 x3 − 1 and its inverse f −1 (x) = (x2 + 1)1/3 . 3