80 CHAPTER 2 Functions and Their Graphs (c) If x 2, = then f 2 2 1 2 2 3 4 ( ) = + + = The point 2, 3 4 ( ) is on the graph of f. (d) If f x 2, ( ) = then x x x x x x x 1 2 2 1 2 2 1 2 4 3 ( ) + + = + = + + = + = − If f x 2, ( ) = then x 3. = − The point 3, 2 ( ) − is on the graph of f. (e) The x-intercepts of the graph of f are the real solutions of the equation f x 0 ( ) = that are in the domain of f. x x x x 1 2 0 1 0 1 + + = + = = − The only real solution of the equation f x x x 1 2 0 ( ) = + + = is x 1, = − so 1− is the only x-intercept. Since f 1 0, ( ) − = the point 1, 0 ( ) − is on the graph of f. f x 2 ( ) = Multiply both sides by x 2. + Distribute. Solve for x. Multiply both sides by x 2. + Subtract 1 from both sides. Now Work PROBLEM 27 NOTE In Example 3, 2− is not in the domain of f, so x 2 + is not zero and we can multiply both sides of the equation by x 2. + j Energy Expended For an individual walking, the energy expended E in terms of speed v can be approximated by E v v v 29 0.0053 ( ) = + where E has units of cal/min/kg and v has units of m/min. (a) Find the energy expended for a speed of v 40 m min. = (b) Find the energy expended for a speed of v 70 m min. = (c) Find the energy expended for a speed of v 100 m min. = (d) Graph the function E E v v , 0 200. ( ) = < ≤ (e) Create a TABLE with TblStart 1 = and Tbl 1. Δ = Which value of v minimizes the energy expended? Source: Ralston, H.J. Int. Z. Angew. Physiol. Einschl. Arbeitsphysiol. (1958) 17: 277. Solution EXAMPLE 4 (a) The energy expended for a walking speed of v 40 = meters per minute is ( ) = + ⋅ ≈ E 40 29 40 0.0053 40 0.94 cal min kg (b) The energy expended for a walking speed of v 70 = meters per minute is E 70 29 70 0.0053 70 0.79 cal min kg ( ) = + ⋅ ≈
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