Tangent Lines and Instantaneous Rates of Change: Derivatives, Exams of Calculus

Download Tangent Lines and Instantaneous Rates of Change: Derivatives and more Exams Calculus in PDF only on Docsity! The Derivative The Tangent Line Let two points on the graph of a function, f(x), be (a, f(a)) and (a + h, f(a + h)). The line passing through these points is called the secant line (see figure 1) and its slope is equal to the average rate of change between the two points. Slope of secant line = average rate of change = ( )( )f a h f a h + − If we let h approach zero, the point (a + h, f(a + h)) will get closer and closer to the point (a, f(a)), as shown in figure 1. This would then result in giving us the instantaneous rate of change where x = a, which is what is called the slope of the tangent line. Therefore, a tangent line is a line that touches the graph of a function at only one point provided that the limit as h approaches zero of the difference quotient exists. Figure 1 This slope and the point (a, f(a)) can then be substituted into the point-slope form of a line to determine the equation of the tangent line. Point-slope form of a line y – y1 = m(x – x1) where m = slope of tangent line and (x1, y1) = the point (a, f(a)) Gerald Manahan SLAC, San Antonio College, 2008 1 Example 1: Find the slope of the tangent line to the graph of f(x) = x2 + 3x – 4 at x = 1. Find the equation of the tangent line. Solution: To determine the slope of the tangent you would begin by finding f(1) and f(1 + h). f(x) = x2 + 3x – 4 f(1) = (1)2 + 3(1) – 4 f(1) = 1 + 3 – 4 f(1) = 0 f(x) = x2 + 3x – 4 f(1 + h) = (1 + h)2 + 3(1 + h) – 4 f(1 + h) = 1 + 2h + h2 + 3 + 3h – 4 f(1 + h) = h2 + 5h Now substitute f(1) and f(1 + h) into the formula for the slope of the tangent line. Slope of tangent line = 0 (1 ) (1)lim h f h f h→ + − 2 0 0 2 0 0 0 (1 ) (1) 5 0lim lim 5lim ( 5)lim lim( 5) 0 5 5 h h h h h f h f h h h h h h h h h h h → → → → → + − + − = + = + = = + = + = Next, substitute the calculated slope and point into the point-slope form of a line. (a, f(a)) = (1, 0) and m = 5 y – y1 = m(x – x1) y – 0 = 5(x – 1) y = 5x – 5 The equation of the tangent line to f(x) at the point (1, 0) is y = 5x – 5. Gerald Manahan SLAC, San Antonio College, 2008 2