Extreme Value Theorem - Intermediate Value Theorem

Example 1: Is there a solution to between and ?
●First, check if the given is continuous.
●Evaluate the functions using the given endpoints, then check
if zero is in between the two computed values.
because , then, according to IVT, the function has at least one solution between
and .

Example 3: Does the function
have a solution on the closed interval [0, 4]?
if one is in between the two computed values.
Since 1 is not between (0) and (4), the IVT fails to hold for the
given function on the closed interval [0, 4] and with

Example 3: Does the function
have a solution on the closed interval [-
2, 2]?
if one is in between the two computed values.
Since , that is , then yes, the function has a solution on [-2, 2].

Extreme Value
Theorem
If is continuous at every point of a closed interval [a, b], then assumes
both an absolute maximum value M and an absolute minimum value m
somewhere in [a, b]. That is, there are numbers 1 and 2 within [a, b],
𝑐 𝑐
such that and , and for every other x in the interval. The highest value of a
function f on a given closed interval is called its absolute maximum value,
and its lowest value is called its absolute minimum value. Collectively,
these values are known as extreme values of on a closed interval

Identify the extreme
values of the function on
the closed interval [-2, 1]
shown in Figure 2.
Solution: On the given interval,
the graph’s lowest value
is 1 which occurs on two values
of
and its highest value is 4 which
occurs when .
Thus, the absolute minimum
value of the function is 1 and
the absolute maximum value is
4.
Example #1

Identify the absolute extreme
values of the
function defined on the interval
[-3, 2].
Solution: First, sketch the graph of
the function on
the given interval, similar to the
one in Figure 3.
Refer to the table of values below.
Example #2
Based on the figure, the function f has an
absolute maximum value of 0 and an absolute
minimum value of -9 on the interval [-3, 2].

Extreme Value Theorem - Intermediate Value Theorem

Example 6: At a school canteen, bottled water is priced P20.00 for
students. At that price, the canteen sells 200 items daily. For every P2.00
increase in price, there are 10 fewer students willing to buy the bottled
water.
a. With this scheme, what will be the revenue function of the school
canteen for the sale of bottled water?
Solution:
a. The revenue function is simply the product of the selling price and
the quantity of items sold. Without the P2.00 increase, the revenue of
the canteen is fixed at P20.00 X 200 = P4,000.00 daily. Let x represent
the number of times that the canteen increases the price of bottled water.
𝑅( ) =
𝑥 𝑠𝑒𝑙𝑙𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 𝑋 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑡𝑒𝑚𝑠 𝑠𝑜𝑙𝑑

Example 6: At a school canteen, bottled water is priced P20.00 for
students. At that price, the canteen sells 200 items daily. For every P2.00
increase in price, there are 10 fewer students willing to buy the bottled
water.
b. You are asked to determine the absolute maximum value of the
function on the interval [0, 10]. To answer the question, you can set up a
table of values for the function and sketch the graph.
If you observe the table above, the maximum
revenue of the canteen will happen after a P2.00 increase is
done 5 times; thus, it will reach a total price of P30.00 If the
canteen decides to increase the price 10 times, the revenue will
be the same as that when sold by the original price.

The Equation of the Tangent Line
The precise definition of a
tangent line relies on the notion
of a secant line. Let C be the
graph of a continuous function
and let P be a point on C. A
secant line to through P is any
line connecting P and another
point Q on C. In the figure on
the right, the line P Q is a
secant line of through P.

If the sequence of secant lines to the
graph of through P approaches one
limiting position (in consideration of
points Q to the left and from the right of
P), then we define this line to be the
tangent line to at P.

Consider the graph of a function whose
graph is given below. Let be a point on
the graph of . Our objective is to find the
equation of the tangent line (TL) to the
graph at the point

• Since the tangent line is the limiting position of
the secant lines as Q approaches P, it follows
that the slope of the tangent line (TL) at the point
P is the limit of the slopes of the secant lines as
x approaches . In symbols
• Finally, since the tangent line passes throughthen
its equation is given by

Example 1:
Find the equation of the tangent line to
𝑥0=2

Example 2:
Find the equation of the tangent line to the graph of
𝑥0=1

Extreme Value Theorem - Intermediate Value Theorem

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