301 2.8 Function Operations and Composition The Difference Quotient As shown previously, the average rate of change of a function on the interval 3a, b4 is given by the slope of the secant line joining the points 1a, ƒ1a22 and 1b, ƒ1b22. See Figure 93(a). We now consider a related idea. Suppose a point P lies on the graph of y = ƒ1x2 as in Figure 93(b), and suppose h is a positive number. If we let 1x, ƒ1x22 denote the coordinates of P and 1x + h, ƒ1x + h22 denote the coordinates of Q, then the line joining P and Q has slope as follows. m= ƒ1x + h2 - ƒ1x2 1x + h2 - x Slope formula m= ƒ1x +h2 −ƒ1x2 h , h 30 Difference quotient This boldface expression is the difference quotient. Figure 93(b) shows the graph of the secant line PQ. As h approaches 0, the slope of this secant line approaches the slope of the line tangent to the curve at P. Important applications of this idea are developed in calculus. EXAMPLE 4 Finding the Difference Quotient Let ƒ1x2 = 2x2 - 3x. Find and simplify the expression for the difference quotient, ƒ1x + h2 - ƒ1x2 h . SOLUTION We use a three-step process. Step 1 In the numerator, find ƒ1x + h2. Replace x in ƒ1x2 with x + h. ƒ1x + h2 = 21x + h22 - 31x + h2 ƒ1x2 = 2x2 - 3x Step 2 Find the entire numerator, ƒ1x + h2 - ƒ1x2. ƒ1x + h2 - ƒ1x2 = 321x + h22 - 31x + h24 - 12x2 - 3x2 Substitute. = 21x2 + 2xh + h22 - 31x + h2 - 12x2 - 3x2 Square x + h. = 2x2 + 4xh + 2h2 - 3x - 3h - 2x2 + 3x Distributive property = 4xh + 2h2 - 3h Combine like terms. From Step 1 Secant line y = f(x) a b f(a) (a, f(a)) y f(b) (b, f(b)) x x x + h P(x, f(x)) Q(x + h, f(x + h)) Secant line y = f(x) f(x) f(x + h) x y h Figure 93 (a) (b) Average rate of change on 3 a, b4 = ƒ1b2 −ƒ1a2 b −a LOOKING AHEAD TO CALCULUS The difference quotient is essential in the definition of the derivative of a function in calculus. The derivative provides a formula, in function form, for finding the slope of the tangent line to the graph of the function at a given point. To illustrate, it is shown in calculus that the derivative of ƒ1x2 = x2 + 3 is given by the function ƒ′1x2 = 2x. Now, ƒ′102 = 2102 = 0, meaning that the slope of the tangent line to ƒ1x2 = x2 + 3 at x = 0 is 0, which implies that the tangent line is horizontal. If you draw this tangent line, you will see that it is the line y = 3, which is indeed a horizontal line.
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