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Recall: Ave Rate of Change & Slope of Secant Line Recall: Ave rate of change & Slope of secant line 2 1 (x,f(x)) 0.5 1.0 1.5 2.0 (1,f(1)) Btwn (1,f(1)) and (x,f(x)), -1 f is changing at an average rate of [f(x)-f(1)]/(x-1) -2 Secant line btwn (1,f(1)) and (x,f(x)) Has slope [f(x)-f(1)]/(x-1) From a; f (a)-3 to x; f (x) f (x) − f (a) Avg r.o.c. = x − a = Slope of secant line Math 101-Calculus 1 (Sklensky) In-Class Work February 20, 2015 1 / 10 Alternatively: Ave rate of change & Slope of secant line 2 h 1 (1+h,f(1+h)) 0.5 1.0 1.5 2.0 (1,f(1)) Btwn (1,f(1)) and (1, f(1+h)), -1 f is changing at an average rate of [f(1+h)-f(1)]/h -2 Secant line btwn (1,f(1)) and (1+h,f(1+h)) Has slope [f(1+h)-f(1)]/(h) From a; f (a)-3 to a; f (a + h) f (a + h) − f (a) Avg r.o.c. = h = Slope of secant line Math 101-Calculus 1 (Sklensky) In-Class Work February 20, 2015 2 / 10 Recall: The derivative function Definition: The derivative function for f (x) is the function: I which gives the slope of the line tangent to f at each point (x; f (x)), if that slope exists. I or equivalently which gives the instantaneous rate that f changes at each point (x; f (x)), if such an average exists. The derivative function is denoted f 0(x) and is read as the derivative of f (x). At every point (x; f (x)) where the function f (x) has a tangent line, the derivative function has a point (x; f 0(x)). Math 101-Calculus 1 (Sklensky) In-Class Work February 20, 2015 3 / 10 The closer b is to a, the closer the slope of the secant line is to the slope of the curve itself at x = a. f (x) − f (a) (slope of tangent line at (a; f (a))) = lim x!a x − a f (a + h) − f (a) = lim h!0 h The smaller the time interval, the closer the average rate of change is to the instantaneous rate of change at t = a. f (t) − f (a) (inst r.o.c. at (a; f (a))) = lim t!a t − a f (a + h) − f (a) = lim h!0 h ) Slope of Curve at x = a = Inst. R. of C. at x = a What We're Really Interested In: I Slope of a Curve (i.e, slope of the line tangent to the curve) I Instantaneous Rate of Change Math 101-Calculus 1 (Sklensky) In-Class Work February 20, 2015 4 / 10 f (x) − f (a) (slope of tangent line at (a; f (a))) = lim x!a x − a f (a + h) − f (a) = lim h!0 h f (t) − f (a) (inst r.o.c. at (a; f (a))) = lim t!a t − a f (a + h) − f (a) = lim h!0 h What We're Really Interested In: I Slope of a Curve (i.e, slope of the line tangent to the curve) The closer b is to a, the closer the slope of the secant line is to the slope of the curve itself at x = a. I Instantaneous Rate of Change The smaller the time interval, the closer the average rate of change is to the instantaneous rate of change at t = a. Math 101-Calculus 1 (Sklensky) In-Class Work February 20, 2015 4 / 10 ) Slope of Curve at x = a = Inst. R. of C. at x = a What We're Really Interested In: I Slope of a Curve (i.e, slope of the line tangent to the curve) The closer b is to a, the closer the slope of the secant line is to the slope of the curve itself at x = a. f (x) − f (a) (slope of tangent line at (a; f (a))) = lim x!a x − a f (a + h) − f (a) = lim h!0 h I Instantaneous Rate of Change The smaller the time interval, the closer the average rate of change is to the instantaneous rate of change at t = a. f (t) − f (a) (inst r.o.c. at (a; f (a))) = lim t!a t − a f (a + h) − f (a) = lim Math 101-Calculus 1 (Sklensky) In-Class Work h!0 h February 20, 2015 4 / 10 ) Slope of Curve at x = a = Inst. R. of C. at x = a Illustrating an example: If f (x) = x2, then we found that f 0(1) = 2 and so the equation of the line tangent to f (x) = 2x − 1. Below is the graph of f (x) = x2 and y = 2x − 1 to illustrate this result: 4 y=x^2 3 y=2x-1 2 (1,1) 1 0.5 1.0 1.5 2.0 -1 Math 101-Calculus 1 (Sklensky) In-Class Work February 20, 2015 5 / 10 In Class Work 1. Use the limit definition to find the slope of the line tangent to f (x) = x2 − 3x at the following points. Remember: If you know a short cut for this process, do NOT use it here. The point of this work is to practice with and understand the limit definition of the slope of the tangent line. (a) x = 0 (b) x = −2 (c) An unspecified (i.e. variable) point x 2. Find the equation of the line tangent to f (x) = x2 − 3x at x = −2. Math 101-Calculus 1 (Sklensky) In-Class Work February 20, 2015 6 / 10 Solutions 1. Use the limit definition to find the slope of the line tangent to f (x) = x2 − 3x at: (a) x = 0 f (0 + h) − f (0) mtan = lim h!0 h (h)2 − 3(h) − 02 − 3(0) = lim h!0 h h2 − 3h = lim h!0 h = lim h − 3 h!0 = −3 Math 101-Calculus 1 (Sklensky) In-Class Work February 20, 2015 7 / 10 Solutions 1. Use the limit definition to find the slope of the line tangent to f (x) = x2 − 3x at the following points. (b) x = −2 f (−2 + h) − f (−2) mtan = lim h!0 h (−2 + h)2 − 3(−2 + h) − 22 − 3(−2) = lim h!0 h 4 − 4h + h2 + 6 − 3h − 4 + 6 = lim h!0 h −4h + h2 − 3h = lim = lim −4 + h − 3 h!0 h h!0 = −7 Math 101-Calculus 1 (Sklensky) In-Class Work February 20, 2015 8 / 10 Solutions 1. Use the limit definition to find the slope of the line tangent to f (x) = x2 − 3x at: (c) An unspecified point x f (x + h) − f (x) mtan = lim h!0 h (x + h)2 − 3(x + h) − x 2 − 3(x) = lim h!0 h x 2 + 2xh + h2 − 3x − 3h − x 2 − 3x = lim h!0 h 2xh + h2 − 3h = lim h!0 h = lim 2x + h − 3 h!0 = 2x − 3 Math 101-Calculus 1 (Sklensky) In-Class Work February 20, 2015 9 / 10 Solutions 2. Find the equation of the line tangent to f (x) = x2 − 3x at x = −2. Need: point and a slope Slope: mtan(−2) = −7 Point: Use point of tangency, − 2; f (−2) = (−2; 4 + 6) = (−2; 10) Thus the tangent line is given by y − 10 = −7(x + 2) =) y = −7x − 4: Math 101-Calculus 1 (Sklensky) In-Class Work February 20, 2015 10 / 10.
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