1. Ch. 2 – Limits and Continuity
2.4 – Rates of Change and Tangent Lines

2. Secant lines pass through a curve at two specific points
Ex: Find the average rate of change of f(x) = x3 – x over the interval [1, 3].
Average rate of change = slope!
The 12 signifies the slope of the secant line through (1, f(1)) and (3, f(3)).
Secant lines pass through a curve at two specific points
Tangent lines pass through a curve at one specific point
Slopes of tangent lines represent INSTANTANEOUS SLOPE at any point on the curve!
Instantaneous slope = rate of change

5. As h approaches zero, we approach the true slope of the tangent line:
To find the slope of the tangent line, consider secant lines that have almost the same slope as the tangent line:
(x+h, f(x+h))
As h approaches zero, we approach the true slope of the tangent line:
(x+h, f(x+h))
(x+h, f(x+h))
(x+h, f(x+h))
(x, f(x))
(x+h, f(x+h)) x
x+h
x+h
x+h
x+h
x+h
HOW TO FIND INSTANTANEOUS SLOPE

6. Ex: Find the slope of the graph of f(x) = x2 at the point (3, 9), and then write an equation for the tangent line at that point.
Use the limit of the difference quotient from the last slide to find slope!
We know the slope is 6 and the line passes thru (3, 9), so…

7. At what x-value will the slope be 15?
Ex: Find the general equation for the instantaneous slope of any point on f(x) = 2x2 – x .
Since we don’t have a specific point, use (x, f(x)) for your point!
At what x-value will the slope be 15?
If the slope is always 4x – 1, then…

8. A normal line is the line perpendicular to a tangent line at a point on the curve.
Ex: Find the equation for the normal line to the curve f(x) = 1/x at the point (2, ½).
That’s the slope of the tangent line…the slope of the normal would be 4 (the opposite reciprocal).
Multiply by 2(2+h)

9. Ex: Does the following curve have a tangent line at x = 0?
The slopes determined from each piece should be equal at x = 0!
Answer: Yes, because the slopes from the left and from the right are both -1.

10. (Multiply by conjugate of numerator)
Ex: Find the derivative of and use it to write an equation of the tangent line through (4, 2).
Now evaluate at x=4 to find slope…
Slope = ¼
The tangent line has a slope of ¼ and passes through (4. 2), so write an equation…
I’ll use point slope form:
(Multiply by conjugate of numerator)