Review: 1) What is a tangent line? 2) What is a secant line? 3) What is a normal line?

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Average Rate of Change What is the formula for the average rate of change? What does it mean?

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Review:1) What is a tangent line?

2) What is a secant line?

3) What is a normal line?

Answers:3. Continuous. x2 + 1 ≠ 06. Continuous It is a composition of functions9. Discontinuous @ x = 012a. Yes f(1) = 1 b. Yes lim = 2 c. No d. No15. f(2) = 018. Yes21a. x = 1 b. Non-removable, jump24a. x = 1,2 b. x = 1 is non-removable x = 2 is removable f(2) = 1

42. Possible Graph

45. x ≈ -0.724 and -1.22151. Consider f(x) = x - e-x, f is continuous, f(0) = -1, and f(1) = 1 - 1/e > .5 By the Intermediate Value Thm, for some c in (0,1), f(c) = 0 and e-c = c

63. Since |x| is continuous, all functions in absolute value are continuous.

Average Rate of Change

What is the formula for the average rate of change?

What does it mean?

Tangents & Secants

A tangent line intersects the curve at exactly one point. The tangent line can describe the rate of change at that point.

A secant line intersects the curve at two or more points. We can use it to determine the slope of the line tangent if the points are infinitesimally close.

Ex 1: Fine the average rate of change forf(x) = x3 - x over the interval [2,4]

Ex 2: Use the points P(23, 150) and Q(45, 350) to computer the average rate of change and slope of the secant line PQ.

Ex 3: Find the slope of the parabola f(x) = x2 at the point P(2,4). Write the equation for the tangent to the parabola at this point.

Slope of a Curve @ a Point

The slope of a curve f(x) at the point (a,f(a)) is

m = lim f(a + h) - f(a) h 0 h

The tangent line to the curve at the point is the line through the point with this slope.

Normal to the curve.

The normal line to a curve at a point is the line perpendicular to the tangent at that point.

f(x) tangent line

normal

Ex 1: Find the slope of the curve at x = a.

a) f(x) = x2 b) f(x) = 1/x

Ex 1: Find the slope of the curve at x = a.

a) f(x) = x2 b) f(x) = 1/x

Ex 3: The equation for free fall at the surface of Jupiter is s = 11.44t2 m with t in seconds. Assume a rock is dropped from the top of a 500 m cliff. Find the speed of the rock at t = 2.

Assignment: p92 # 1-33odd

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