<p> 1</p><p>Section 2.1: The Derivative and the Tangent Line Problem</p><p>Practice HW from Larson Textbook (not to hand in) p. 87 # 1, 5-23 odd, 59-62</p><p>Tangent Lines</p><p>Recall that a tangent line to a circle is a line that touches the circle only once.</p><p>If we magnify the circle around the point P</p><p> we see that slope of the slope of the tangent line very closely resembles the slope of the circle at P. For functions, we can define a similar interpretation of the tangent line slope.</p><p>Definition: The tangent line to a function at a point P is the line that best describes the slope of the graph of the function at a point P. We define the slope of the tangent line to be equal to the slope of the curve at the point P.</p><p>Examples of Tangent Lines: 2 3</p><p>Consider the following graph:</p><p>Slope of Secant line between the points = (x, f(x)) and (x+h, f(x+h))</p><p>As h→0, the slope of the secant lines approach that of the tangent line of f at x = a.</p><p>Slope of tangent line = m = of f at (x, f(x)) 4</p><p>Definition: Given a function y  f (x) , the derivative of f, denoted by f  , is the function defined by </p><p> f (x  h)  f (x) f (x)  lim , h0 h provided the limit exists.</p><p>Facts 1. For a function y  f (x) at a point x = a, f (a) gives the slope of the tangent line to the graph of f at the point (a, f (a)) . 2. f (a) represents the instantaneous rate of change of f at x = a, for example, instantaneous velocity.</p><p>Example 1: Use the definition of the derivative to find the derivative of the function f (x)  3x  2 .</p><p>Solution: </p><p>█ Example 2: Use the definition of the derivative to find the derivative of the function f (x)  2x2  x 1. 5</p><p>Solution: </p><p>█ Example 3: Find the equation of the tangent line to the curve f (x)  x3  3x 2  4 at the point (1, 2).</p><p>Solution: 6</p><p>█ 7</p><p>Example 4: Use the definition of the derivative to find f (x) if f (x)  1 x .</p><p>Solution: Using the limit definition of the derivative, we see that</p><p> f (x  h)  f (x) f (x)  lim h0 h</p><p>1 x  h  1 x  f (x)  1 x     lim  Note that  h0 h  f (x  h)  1 x  h </p><p>( 1 x  h  1 x) ( 1 x  h  1 x) Multiply numerator and denominator  lim   h0 h ( 1 x  h  1 x)  by the conjugate of the numerator </p><p>(1 x  h)  1 x  h 1 x  1 x  h 1 x  (1 x) FOIL the numerator and  lim   h0 h ( 1 x  h  1 x)  multiply denominator </p><p>1 x  h 1 x  lim (Simplify and distribute - sign to 1 x) h0 h ( 1 x  h  1 x)</p><p> h  lim (Simplify) h0 h ( 1 x  h  1 x)</p><p>1  lim (Simplify by canceling h terms) h0 1 x  h  1 x</p><p>1  (Substitute h  0 to evaluate the limit) 1 x  0  1 x</p><p>1  (Simplify) 1 x  1 x</p><p>1  (Simplify by combining like terms) 2 1 x</p><p>█ 8</p><p>Example 5: Find the derivative using the definition and use the result to find the equation of the 1 line tangent to the graph of f (x)  at the point (0, -1). x 1</p><p>Solution: To find the equation of any line, including a tangent line, we need to know the line’s slope and a point on the line. Since we already have a point on line, we must find the tangent line’s slope, which is found using the derivative. Using the limit definition of the derivative, we see that</p><p> f (x  h)  f (x) f (x)  lim h0 h</p><p>1 1   1 1   lim x  h 1 x 1  Note that f (x)  , f (x  h)   h0 h  x 1 x  h 1</p><p>1 [x 1] 1 [x  h 1]  (x  h 1) [x 1] (x 1) [x  h 1]  lim The common demonimator is (x  h 1)(x 1)  h0 h</p><p>(x 1) (x  h 1)  (x  h 1)(x 1) (x 1)(x  h 1)  lim Rewrite fractions in terms of common denominator h0 h</p><p>(x 1)  (x  h 1) (x  h 1)(x 1)  lim (Combine fractions over common denominator) h0 h</p><p>(x 1  x  h 1) (x  h 1)(x 1)  lim (Distribute - sign) h0 h</p><p> h (x  h 1)(x 1)  lim (Simplify) h0 h  h (x  h 1)(x 1) h  lim (Write h as ) h0 h 1 1  h 1  lim  (Take reciprocal of denominator and multiply) h0 (x  h 1)(x 1) h</p><p>1  lim (Cancel h terms) h0 (x  h 1)(x 1) 1 1 1     (Simplify to get the derivative ) (x  0 1)(x 1) (x  0 1)(x 1) (x 1)2 Continued on next page 9</p><p>1 Using f (x)   , we can now find the slope at the give point (0, -1). (x 1)2</p><p>Slope at the point 1 1 1 (0,1)  m  f (0)        1. 2 2 1 (x  0) (0 1) (1)</p><p>Using m  1, we see that from the slope intercept equation y  mx  b that y  x  b</p><p>To find b, use the fact that at the point (0, -1), x  0 and y  1. Thus</p><p>1  0  b giving b  1. Thus, the equation of the tangent line is:</p><p> y  x 1. 10</p><p>Differentiability and Points where the Derivative Does Not Exist</p><p>Note: The derivative of a function may not exist a point.</p><p>Fact: If a function is not continuous at a point, its derivative does not exist at that point.</p><p>1 For, example, f (x)  is not continuous at x = 1. This implies that f (1) will not x 1 1 exist. Note that f (x)   , which computationally says f (1) does not exist. (x 1)2</p><p>Note! However, a function may still be continuous at a point but the derivative may still not exist. 11</p><p>Example 10: Use the definition of the derivative to demonstrate that f (0) does not exist for the function f (x)  x .</p><p>█ 12 13</p><p>Fact: In general, a function that displays any of these characteristics at a point is not differentiable at that point</p><p>Cases</p><p>1. The function is not continuous at a point – it has a jump, break, or hole in the graph at that point. </p><p>2. The function has a sharp point or corner at a point.</p><p>3. The function has a vertical tangent at a point.</p><p>14</p><p>Example 11: Determine the point)s on the following graph where the derivative does not exist. Give a short reason for your answer.</p><p>Solution:</p><p>█</p>