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which can be morphed algebraically into y = (x + 5)1/3 Replacing y by h 1 (x), we get h 1 (x) = (x + 5)1/3 4 Let s examine each situation by imagining what happens when we try to input real numbers to the original relations, and working from there: Problem 1 and its solution: The relation and its inverse are f (x) = 2x + 4 and f 1(x) = x / 2 2 We can input any real number into either of these relations, and we always get a meaningful result Therefore, the domains of f and f 1 both span the entire set of real numbers The range of f 1 is the same as the domain of f, and the range of f is the same as the domain of f 1 Therefore, the ranges of f and f 1 both include all real numbers Problem 2 and its solution: The relation and its inverse are g (x) = x 2 4x + 4 and g 1(x) = 2 x1/2 We can input any real number we choose into g, and we always get a meaningful result However, that result is always nonnegative, because it s the square of the real-number quantity (x 2) It s possible for the output of g to equal 0; that happens when the input is 2 The range of g is therefore the set of all nonnegative reals The domain of g 1 is identical to the range of g, so the domain of g 1 spans the set of all nonnegative reals The range of g 1 is the same as the domain of g, which is the set of all reals Problem 3 and its solution: The relation and its inverse are h (x) = x 3 5 and h 1(x) = (x + 5)1/3 We can input any real number into h, and we always get a real-number output value Therefore, the domain of h spans the entire set of reals The same is true of the domain.

12

Not only does the slope of the curves change with temperature, but it Any horizontal line drawn across Fig. 9.4 also changes with load WH. will present the conditions required for temperature control. Doubling of the load at any given temperature requires the manipulated variable WC to be much more than doubled. In practice, the overall heat transfer coefficient aIso varies with the flow rates, which improves the controllability somewhat. Although the film coefficient on each side of the heat transfer surface varies at about the 0.8 power of the fluid velocity, for simplification it will be assumed that the relationship is linear. Furthermore the reciprocal of the overall heat transfer coefficient mill be assumed to be the sum of the reciprocals of the individual film coefficients: 1 1 1 -= u W&H + W&c The terms kH and kc are the flow indices of their respective heat transfer coefficients. Combining with Eq. (9.11) yields:

of h 1; the relation is defined for all possible real-number input values The range of h 1 is the same as the domain of h, and the range of h is the same as the domain of h 1 Therefore, the ranges of h and h 1 both span the set of all reals 5 The equation of our relation, as given in the problem, is x 2 /4 y 2 /9 = 1 Let s use algebra to morph this equation so it s stated with y all by itself on the left-hand side and an expression containing only the variable x on the right-hand side If we multiply through by 36, we get 9x 2 4y 2 = 36 which can be rewritten as 4y 2 = 36 9x 2 Multiplying through by 1, we get 4y 2 = 9x 2 36 When we divide through by 4, we obtain y 2 = 9x 2 /4 9 Taking the complete square root of both sides yields y = (9x 2 /4 9)1/2 Using relation notation to express this equation and name the relation f, we get f (x) = (9x 2 /4 9)1/2 That s the relation! Now let s find its inverse We begin by restating the relation with y on the left instead of f (x), so we have y = (9x 2 /4 9)1/2 Swapping the names of the variables gives us x = (9y 2 /4 9)1/2 When we square both sides, we obtain x 2 = 9y 2 /4 9