Download MATH 150 SUPPLEMENTAL NOTES 6 TANGENT LINES AND THE

MATH 150 SUPPLEMENTAL NOTES 6
TANGENT LINES AND THE DEFINITION OF THE DERIVATIVE
SLOPE OF THE SECANT LINE AND THE DEFINITION OF THE DERIVATIVE
Consider the function f (x) = x 2 - 4x + 4 and its graph.
(See figure 1) We would like to determine the slope of the
tangent line at the point (1, 1). The way we will do this is by
finding successive slopes of secant lines that go through that
point. Remember that a secant line is a line that intersects the
curve twice.
figure 1
SECANT LINE
POINTS ON THE CURVE
SLOPE
Secant line 1
(1, 1) → (3, 1)
m=0
Secant line 2
(1, 1) → (2.5, .25)
m = -.5
Secant line 3
(1, 1) → (2, 0)
m = -1
Secant line 4
(1, 1) → (1.5, .25)
m = -1.5
If I plot these secant lines on the graph, you should notice a trend.
What do you notice? - As the second point gets closer to the point
(1, 1), the slope is becoming steeper (negatively). What do you
think the limit of the slopes will be? (It will be the slope of the
tangent line.)
Suppose h is the difference between x and x0, such that x0 = x + h,
then the slope of the secant line at the point x0 is the following:
(For those of you that had me for pre-calculus, you should recognize this formula as the difference quotient, and it is also the average rate of change. Yes, it
does come back to haunt you!! J )
So, for the above function:
This formula simplifies down to the following: m = 2x + h - 4. (Prove this by performing the algebra.) Remember, that we want to determine what the slope
of the tangent line is, so by letting h get smaller and smaller, (i.e. go to zero), we might be able to determine the slope. To do this, we must find the limit of
m as h goes to zero.
(HINT: This expression is f ' (x).)
So, f ' (x) = 2x - 4. This is the equation to find the slope at any point on that curve. So when x = 1, m = -2. If x = 6, then m = 8. Pretty slick isn't it.
NOTE: Here is some new notation for you to know: f ' (x) = y' = dy/dx. They all represent the derivative of y with respect to x. Recall that m = (y2 - y1)/(x2 x1). It can also be thought of as the change in y over the change in x. m = Δ y/Δ x.
EXAMPLE 1:
SOLUTION:
Using the definition of the derivative, determine the equation for the slope of the function f (x) = x 3 - 3x 2 +
3x -1, and then determine the equation of the tangent line at the point x = 2.
When x = 2, f (2) = 1, so our ordered pair is (2,1). Recall that m = f ' (x), so the slope of the tangent line at x = 2 is f ' (2) = 3.
y - 1 = 3(x - 2) → y - 1 = 3x - 6 → y = 3x - 5 (This is the tangent line of the curve at x = 2.)
EXAMPLE 2:
Using the definition of the derivative, determine the equation for the slope for the following function.
(A) Determine the equation of the tangent line at the point x = 3. (B) What is the slope of the tangent line at
the point x = -2? (C)What can you conclude about this point?
SOLUTION:
When x = -2, f' (-2) is undefined. If you look at the original function, you should first realize that x = -2 is not in the domain of f (x). In
fact, x = -2 is a vertical asymptote of this function. f ' (-2) tells us that slope does not exist at x = -2. If the slope does not exist, then it must
be a vertical line. Here the graph is becoming vertical in nature. So, when the first derivative does not exist, it could mean that we have a
vertical asymptote.
EXAMPLE 3:
Using the definition of the derivative, determine the equation for the slope for the following function.
(A) Determine the equation of the tangent line at the point x = 2. (B) Determine the slope of the tangent line
at the point x = -2. (C) What can you conclude about this point?
SOLUTION:
When x = -2, f ' (-2) is undefined. Again, what is happening here is the fact that the tangent line at the endpoint of the domain is vertical.
To see this, use your graphics calculator to see this.
DIFFERENTIABLE ON AN INTERVAL - ONE-SIDED DERIVATIVES
FACT:
A function y = f (x) is differentiable on an open interval (finite or infinite) if it has a derivative at each point on the
interval.
FACT:
A function is differentiable on a closed interval [a, b], if it is differentiable on the interior (a, b) and if the limits
exists at the endpoints.
Here is an example of a function that is continuous on the interval (-∞ , ∞ ), but does not have a derivative at a specific point. Consider the
function f (x) = | x + 3 |. This function is differentiable on the interval (-∞ , -3) and (-3, ∞ ), but does not have a derivative at x = -3. Here is
the proof of this assertion.
RIGHT-HAND LIMIT
Since we are approaching zero from the right-hand side, then h is positive and | h | = h.
LEFT-HAND LIMIT
Since we are approaching zero from the left-hand side, then h is negative and | h | = -h.
Since the limits approach different values as h goes to zero, then the derivative does not exist at x = -3.
Now the question should be this - when does a function not have a derivative at a point? So let us investigate this question.
WHEN DOES A FUNCTION NOT HAVE A DERIVATIVE AT A POINT?
As the text states, there are four types of functions that have points where the derivative does not exist. Let us look at each one up close and
personal.
A CORNER
Consider the graph of the function
f (x) = | x - 2 |.
Notice that the graph has a corner. So the question should be
this, what is the right-hand and left-hand limits at the point x =
2?
QUESTION 1: What is the right-hand limit?
QUESTION 2: What is the left-hand limit?
Since, both the right- and left-hand limits do not equal, then the derivative does not exist at the point x = 2.
A CUSP
Consider the graph of the function
f (x) = - x 2/5.
Notice that this graph comes to a cusp at the point (0, 0). So let
us investigate the right- and left-hand limits at this point.
QUESTION 1: What is the limit doing as h → 0 -?
Let us look at a sequence of secant lines to discover what this
limit is doing.
The secant line y = (1/27) x.
The secant line y = (1/8) x.
The secant line y = (1/4) x.
Do you see anything happening here, as the secant line's slope
becomes steeper and steeper? The slope of the secant line is
approaching ∞ as h → 0 -. Therefore, the limit does not exist.
QUESTION 2: What is the limit doing as h → 0 +?
Again, let us look at the secant lines.
The secant line y = - (1/27) x.
The secant line y = - (1/8) x.
The secant line y = - (1/4) x.
Do you see anything happening here, as the secant line's slope
becomes steeper and steeper? The slope of the secant line is
approaching ∞ as h → 0 +. Therefore, the limit does not exist.
From these conclusions, we can conclude that a graph that has a
cusp in it is not differentiable at the point where the cusp
occurs.
A VERTICAL LINE
Consider the graph of the function
f (x) = x 1/ 3.
Let us determine if there will be a derivative at the point (0, 0)
on this graph. We will do this by looking at a sequence of
slopes of secant lines.
QUESTION 1: What does the limit approach as h → 0 -?
The secant line y = 0.25 x.
We will investigate this question by looking at the following
graphs.
The secant line y = 0.5 x.
The secant line y = x.
You should notice that as a point Q approaches the point (0, 0), the secant line begins to take on the characteristics of a vertical line. Hence
the limit as h → 0 - goes to ∞ . If we look at the limit as h → 0 +, the secant lines will indicate that this limit is approaching -∞ . Therefore,
the derivative does not exist at the point (0, 0).
A DISCONTINUITY
Any rational function that has a point of discontinuity - removable or asymptotic - will satisfy this type. One function that you might be
familiar with is y = 1/ x. This graph has vertical asymptote at x = 0. Therefore, the derivative will not exist at the point when x = 0.
Now that we know where functions cannot have derivatives, we now can make a conclusion about the relationship between
differentiability and continuity.
FACT:
If f has a derivative at x = c, then f is continuous at x = c.
FACT:
INTERMEDIATE VALUE PROPERTY OF DERIVATIVES
If a and b are any two points in an interval on which f is differentiable, then f ' takes on every value between f ' (a) and
f ' (b).
In this set of supplemental notes, I have related the limit of the slopes of secant lines to the definition of the derivative.
To find the limit of the slopes, use the difference quotient (a.k.a. the average rate of change) to find the generic slope of
the secant line, then find the limit of this expression as h approaches zero. This will give us the formula for the slope of
the tangent line, which is also the first derivative. So remember that the first derivative gives you the slope of the
tangent line to the curve at any point on the curve. I worked several examples of finding the derivative using the
definition. Then the topic changed to when does a function not have a derivative at a point. There are four types of
functions that do not have derivatives at a point. I have illustrated each type with graphs of the function, and a sequence
of secant lines. This topic lead to an important fact. If a function is differentiable at a point, then it is continuous at the
point. Work through the examples using the definition of the derivative, and make sure you understand why certain
functions do not have derivatives at specific points. If you have any questions on any of the material contained in this
set of supplemental notes, please feel free to contact me.
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