Math 131. The Derivative and the Tangent Line Problem Larson Section 2.1

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1 Math 131. The Derivative and the Tangent Line Problem Larson Section.1 From precalculus, the secant line through the two points (c, f(c)) and (c +, f(c + )) is given by m sec = rise f(c + ) f(c) f(c + ) f(c) = = run c + c See the diagram below The tangent line, if it exists, to the graph of f at (c, f(c)) is the line whose slope matches the slope of the graph of f at (c, f(c)), this slope will be defined as the limit of the slope of the secant lines through as (c, f(c)) and (c +, f(c + )) as 0. Example 1. Consider the function f(x) = 1/x. (a) Sketch the graph of f along with its secant lines through the points (c, f(c)) and (c +, f(c + )) where c = and (i) = 1.5; (ii) = 1; (iii) = 0.5. The sketch the tangent line to the graph. Solution: The graph of f(x) is given in blue: when = 1.5 the secant line goes through (, f()) and (0.5, f(0.5)); when = 1 the secant line goes through (, f()) and (1, f(1)); when =.5 the secant line goes through (, f()) and (1.75, f(1.75)); the secant lines are drawn in red, and the tangent line is drawn in black.

2 y 1 1 x Notice that the secant lines in the picture get closer and closer to the tangent line, and this is how a tangent line of a function is defined. Definition of Tangent Line. Suppose f is defined on an open interval containing c. If the limit f(c + ) f(c) lim = m exists, then the line passing through (c, f(c)) with slope m is the tangent line to the graph of f at the point (c, f(c)). Example. Find the slope and equation of the tangent line to the graph of f(x) = 1/x at the point (, 1/). Solution: The slope is given by f(c + ) f(c) m (+) ()(+) 1 ( + ) = 1 4 ((+) The tangent line has slope m = 1/4 and goes through (, 1/) and so it has point-slope equation y 1 = 1 4 (x ) and then y = 1 x + 1 is the equation in slope-intercept 4 form. (This agrees with the graph of the tangent line in Example 1 because that line has y-intercept (0, 1) and slope of 1/4 since it goes through the points (0, 1) and (, 1/)). A crucial operation in calculus is differentiation, it provides the instantaneous rate of change of a function, and is defined as follows. Definition of Derivative. The derivative of f at x is defined by f (x) f(x + ) f(x)

3 provided the limit exists; and for all x for which the limit exists, f (x) is a function of x. Notice that the derivative function f (x) provides the slope of the tangent line to the graph of f at the point (x, f(x)). For a function y = f(x), you will see various notations for the derivative, including: f (x), dy dx, y, d dx [f(x)] The following is an example of finding a derivative using the limit process. Example 3. (a) Let f(x) = x x + 1. Use the definition of derivative to find f (x). (b) Find the line tangent to the graph at the point (, f()). Then graph both f and this tangent line. Solution: (a) For f(x) = x x+1, the derivative is found using the definition as follows. f f(x + ) f(x) (x) (x + ) (x + ) + 1 (x x + 1) (x + x + () ) x + 1 x + x 1 ()(x + ) x + = x. (b) Now, f() = 1 and at the point (, 1), the tangent line has slope Then the tangent line is m = f () = () =. y 1 = (x ) that is y = x 3 The graph of f(x) = x x + 1 and y = x 3 are given below: 4 y 4 4 x 4

4 There is an alternate form of the definition of derivative of f at c. That is, the original definition is f f(c + ) f(c) (c). If we let x = c +, then = x c and as 0, x c, so the the limit becomes provided the limit exists; see the figure below. f (c) f(x) f(c) x c This alternate definition can be used for finding derivatives using the limit process, for example Example 4. Let f(x) = x 3 5x + 4. Use the alternate definition of the derivative to find f (c). Solution: Given f(x) = x 3 5x + 4 we compute f (c) f(x) f(c) x c x 3 5x + 4 (c 3 5c + 4) x c (x 3 c 3 ) 5(x c) x c (x c)(x + cx + c ) 5(x c) x c (x c)[(x + cx + c ) 5] x c (x + cx + x ) 5 = 6c 5 The alternate definition ensures that differentiability implies continuity.

5 Theorem. Suppose f is differentiable at x = c. Then f is continuous at x = c. Proof. To take a derivative, f must be defined at x = c, also lim[f(x) f(c)] = [ f(x) f(c) lim x c = f(x) f(c) lim x c = f (c) 0 = 0 ] (x c) lim (x c) Now lim f(x) f(c) = 0, means lim f(x) = f(c), and this ensures continuity. However, a function can be continuous but not differentiable. For example, find the derivative of f(x) = x 3 at x = 3 if it exists. For this, f (3) x 3 x x 3 if it exists, but it does not since lim x 3 + x 3 x y x 3 x 3 x 3 = 1, but lim x 3 x 3 x 3 = f(x) = x x Graphically, you can recognize the point of nondifferentiability by the cusp or sharp point on the graph when x = 3. The slope of the graph to the right of 3 is 1, and to the left of 3 is 1, and that is why the limit in the definition of the derivative did not exist. In general, a function is not differentiable where it is not continuous, and typical ways in which continuous functions can fail to be differentiable are where its graph has a cusp, or vertical tangent.

6 Example 5. A function f is graphed to the right below. (a) Identify all x for which f (x) does not exist. For each of those values of x explain why f (x) does not exist. (b) What is the approximate value of f (3)? Explain your answer. (c) What is the approximate value of f ( )? Explain your answer. (d) What is the approximate value of f ( 5)? Explain your answer. Solution: A graph with the tangent lines at relevant points is given below right. (a) f does not exist at x = 1 and x = because the graphs have cusps (the right-hand and left-hand derivatives are different); f does not exist at x = 0 because there is a vertical tangent there. (b) f (3) = 0 because the graph has a horizontal tangent line there. (c) f ( ) = 1 because the tangent line will have slope 1 there. (d) f ( 5) = 1 because the slope of the tangent line there is 1. Graphs of the tangent lines are given.

7 The following are further examples, related to derivatives. Example 6. The limit 5 x lim x 1 x 1 represents f (c) for a function f(x) and a number c. Find f(x) and c. Solution: The alternate definition of derivative is f f(x) f(c) (c), and so we x c try c = 1 (because this is the number x approaches), and then the numerator suggests, f(x) f(c) = 5 x so we let f(x) = 5 x + 3 and check that f(1) = = (5)() = 10 and so with these assignments we have f (c) f(x) f(c) x c 5 x x 1 x 1 = 5 x x 1 x 1 which agrees with the requested information. This means f(x) = 5 x + 3 and c = 1. Example 7. (a) Find the derivative of f(x) = 5 x 3 using the limit process. (b) Use your result in (a) to find the tangent line to the graph of f at the point (4, f(4)). Write your answer in slope-intercept form. Solution: (a) Using the definition, f (x) is computed as follows: f f(x + ) f(x) (x) 5 5 (x+) 3 x 3 5(x 3) 5(x+ 3) (x+ 3)(x 3) 5() (x+ 3)(x 3) 5 (x + 3)(x 3) 5 = (x 3) 5() (x + 3)(x 3) (b) When x = 4, f (4) = 5 5 = 5 and f(4) = = 5. The tangent line is then (4 3) 4 3 y 5 = 5(x 4) and so y = 5x + 5. A graph is given below for reference 1

8 10 y x Example 8. (a) Use the definition of derivative to find the derivative of f(x) = x + 5. (b) Find the equation of the tangent line to the graph at the point (, 3). Solution: (a) Using the definition, the derivative of f is computed as follows: f f(x + ) f(x) (x) (x + ) + 5 x + 5 ( (x + ) + 5 x + 5)( (x + ) x + 5) ()( (x + ) x + 5) (x + ) + 5 (x + 5) ()( (x + ) x + 5) () ()( (x + ) x + 5) = (x + ) x + 5 x + 5 = 1 x + 5 (b) When x =, f 1 () = = 1, so the tangent line has slope 1/3 at the point () (, 3). The equation of the tangent line is then y 3 = 1 (x ) which simplifies to 3 y = 1x + 7. The graph is as follows 3 3

9 Example 9. The graph of the function f(x) = x + 10x 1 along with two tangent lines is given below y x (a) Factor the expression for f(x) to verify the x-intercepts are (3, 0) and (7, 0). (b) Use the limit process to find f (x). (c) Find the two lines through the point (5, 8) that are tangent to the graph of f. Solution: (a) f(x) = x +10x 1 = (x 10x+1) = (x 3)(x 7), the x-intercepts occur when f(x) = 0, and so when x = 3 and x = 7. Thus the x-intercepts are (3, 0) and (7, 0).

10 (b) For f(x) = x + 10x 1, the derivative is found using the definition as follows. f f(x + ) f(x) (x) (x + ) + 10(x + ) 1 ( x + 10x 1) ( x x () ) + 10x x 10x + 1 ()( x ) x = x (c) The tangent lines, per the graph, are tangent at the points (3, 0) and (7, 0), so we can find the equations using the slope from the derivative (there is a way to find the points of tangency purely analytically, but this time, we will use the graph since it is there, and verify the lines are correct at the end). The line with positive slope has slope m = (3) + 10 = 4, the line then has equation y 0 = 4(x 3) or y = 4x 1. Note, when x = 5, y = (4)(5) 1 = 8, so the line goes through (5, 8). The line with negative slope has slope m = (7) + 10 = 4, it thus has equation y 0 = 4(x 7) or y = 4x + 8. Note, when x = 5, y = ( 4)(8) + 8 = 8, and so the line goes through the point (5, 8):

11 Example 10. Let f(x) = 3x 3 7x + 7. Use a limit process to find the derivative f (x). Solution: Given f(x) = 3x 3 7x + 7 we compute f f(x + ) f(x) (x) ( 3(x + ) 3 7x + 7 (3x 3 7x + 7) 3(x 3 + 3x () + 3x() + () 3 ) 7(x + x + () ) + 7 (3x 3 7x + 7) 3x 3 + 9x () + 9x() + 3() 3 7x 14x 7() x 3 + 7x 7 9x () + 9x() + 3() 3 14x 7() )(9x + 9x() + 3() 14x 7()) 9x + 9x() + 3() 14x 7() = 9x 14x