Download Chapter 3: Applications of Differentiation

Chapter 3: Applications
of Differentiation
3.1 Extrema on an Interval
 If
a function is continuous on a
closed interval [a, b], then the
function has both a minimum
and a maximum between (or
including) the points a and b.
The Extreme Value Theorem
 Relative
extrema (minimums
and maximums) occur if there is
an open interval (rather than
closed) on which there is a
maximum value or a minimum
value.
 FYI: These may also be referred
to as local maxima or minima.
Definition of Relative Extrema
 Formal
definition: Let f be defined
at c. If f’(c) = 0 or if f is not
differentiable at c, then c is a
critical number of f.
 What you’ll do: Set f’(x) = 0.
Solve for x. These will be critical
numbers on the function. If you
can’t take the derivative of a
function at a certain point, that is
also a critical number.
Definition of a Critical Number
 If
a function has a relative
minimum or a relative maximum
at x = c, then c is a critical
number of the function.
Relative Extrema and Critical
Numbers
 Finding
Extrema on a Closed
Interval
 𝑓 𝑥 = 3𝑥 4 − 4𝑥 3 on the interval
[-1, 2].
 Step 1: Take the derivative
 𝑓 ′ 𝑥 = 12𝑥 3 − 12𝑥 2
 Step 2: When is this equal to 0?
 12𝑥 2 𝑥 − 1 = 0; 𝑥 = 0; 𝑥 = 1
EXAMPLES
 Step
3: Evaluate the function
using the points -1, 0, 1, and 2
(the endpoints and the zeros).
 (-1, 7), (0, 0), (1, -1), (2, 16)
 Step 4: On this function, the
minimum is (1, -1) and the
maximum is (2, 16)
Examples
 Find
the extrema of
2
3
𝑥 = 2𝑥 − 3𝑥 on the interval
[-1, 3]
𝑓
 Step
 Step
1: 𝑓 ′ 𝑥 = 2 − 2𝑥
2: Set f’(x) = 0
−1
3
−1
3
=0; x = 1 and 0 (f’(x) does
not exist at x = 0)
 2 − 2𝑥
Examples
 Step
3: Find f(-1), f(0), f(1),
and f(3)…
 (-1, -5), (0, 0), (1, -1), (3, -.24)
 Step 4: (-1, -5) is a minimum;
(0, 0) is a maximum
Examples
 Find
the extrema of
f(x) = 2sinx – cos2x on the
interval [0, 2π]
 Step 1: f’(x) = 2cosx + 2sin2x
 Step 2: 2cosx + 2sin2x = 0
 Rewrite sin2x as 2sinxcosx, so
 2cosx + 4sinxcosx = 0
 2cosx(1 + 2sinx) = 0
 x = π/2 and 3π/2 and 7π/6 and 11π/6
Examples
 Step
3: f(0) = -1; f(π/2) = 3;
f(7π/6) = -3/2; f(3π/2) = -1; f(11π/6) =
-3/2; f(2π) = -1
 Step 4: Minimums at (7π/6, -3/2) and
(11π/6, -3/2) and maximum at (π/2, 3)
Examples
 1.
𝑓 𝑥 =3 −
𝑥 𝑜𝑛 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 −1, 2
 2.
𝑓 𝑥 =
2𝑥+5
3
on the interval
[0, 5]
2
 3. 𝑓 𝑥 = 𝑥 −
2𝑥 𝑜𝑛 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 [0, 4]
Examples for You to Try
 Page
169
 21 – 35 odd
Classwork/Homework