10.3.1 Exercises 10.33: An ordinary function graph combines domain and range information in a single picture: we plot the ordered pair (x, f(x)). several variables we quickly run out of pictures...
10.3.1 Exercises 10.33: An ordinary function graph combines domain and range information in a single picture: we plot the ordered pair (x, f(x)). several variables we quickly run out of pictures we can draw, but there is simple alternative, which we illustrate first with an ordinary function from R to R. (This is exactly a domain-range picture, introduced in Section 1.3.) Take f(x) 2. Draw the domain of the function as a single vertical line (fair: the domain is R). Draw the codomain (output space) as a single vertical line to the right of the first. We know f(2) - 4, and we show this by an arrow with tail at 2 in the domain space and with head at 4 in the codomain. Draw more relevant arrows; how does the picture show f is not injective? Not surjective (onto)? For functions of Limits may be included. Not only is f im is surely an e-6 statement. Associated with some e >0 there is an interval (range space, please) into which values of the function must fall. a δ > 0, giving a punctured neighborhood N about 2 about 4 Draw it. We can find in the domain space so that for any z in N, f(x) is in the e-neig In the picture, any arrow with its tail in the 6-neighborhood
10.3.1 Exercises 10.33: An ordinary function graph combines domain and range information in a single picture: we plot the ordered pair (x, f(x)). several variables we quickly run out of pictures we can draw, but there is simple alternative, which we illustrate first with an ordinary function from R to R. (This is exactly a domain-range picture, introduced in Section 1.3.) Take f(x) 2. Draw the domain of the function as a single vertical line (fair: the domain is R). Draw the codomain (output space) as a single vertical line to the right of the first. We know f(2) - 4, and we show this by an arrow with tail at 2 in the domain space and with head at 4 in the codomain. Draw more relevant arrows; how does the picture show f is not injective? Not surjective (onto)? For functions of Limits may be included. Not only is f im is surely an e-6 statement. Associated with some e >0 there is an interval (range space, please) into which values of the function must fall. a δ > 0, giving a punctured neighborhood N about 2 about 4 Draw it. We can find in the domain space so that for any z in N, f(x) is in the e-neig In the picture, any arrow with its tail in the 6-neighborhood