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(1) Use the limit definition of derivative to find f â€² (3) where f (x) = x2 âˆ’ 1. Then check your
answer by using differentiation rules for to get f â€² (x) and then let x = 3.
(2) Name each differentiation rule you use to find the derivative of
x2 sin(2x)
(3) Find an equation of the tangent at the point (0, 1) to the graph of y = cos x. Check it
with desmos.
(4) (a) Find interval of which f (x) = x2 + 2x is decreasing and interval on which it is
increasing. (b) Find the concavity intervals of g(x) = 31 x3 âˆ’ 1.
(5) Multiple choice.
(i) The expression sin ex is
(a) the product of sine and the exponential
(b) does not make sense
(c) the composition of the outer function sine and inner the exponential.
(ii) The Mean Value Theorem on the interval [a, b] for f is a theorem about the existence
of c between a and b such that
(a) f (c) = a + b.
(b) f (c) = f (a) or f (c) = f (b)
(c) the tangent at (c, f (c)) is parallel to the secant line passing through (a, f (a)) and
(b, f (b)).
A critical number is a candidate for a local minimum or maximum. However, not all critical
numbers give rise to a local extremum. In this lesson we illustrate the conditions that ensure
when a critical number in fact gives a local extremum.
This is perhaps one of the more time consuming lessons in the course. All activities:
watching and interacting with the video lesson, reading notes and textbook, contributing on the forum, and writing and submitting the solutions may perhaps take even
8 hours or so. Please allocate enough time. Have I seen you initiating a thread on
the forum? There is no better time to get more involved. The assignments are now
getting quite laborious.
1. Critical numbers are necessary but not enough for local extrema
Recall that a function f has a local minimum at x = a, if its values near x = a are not less
than f (a).
In lesson 14 we saw that the search for local minima can be restricted to critical numbers but
not all critical numbers give rise to a maximum or minimum.
For example, The function f (x) = (x âˆ’ 1)3 + 2 has a critical number x = 1 (f â€² (1) = 0) and
yet it does not have a local minimum nor a maximum. This is because to the left of x = c, f
increases and also f increases to the right of x = c.
Thus the mere existence of a critical number does not ensure that a function has a minimum
there. In order for a function f to have a local minimum at a critical number x = c, f must
change its monotonicity at x = c from decreasing to the left of x = c to increasing to the right
of x = c.
Similarly, in order for a function f to have a local maximum at a critical number x = c, f
must change its monotonicity at x = c from increasing to the left of x = c to decreasing to the
right of x = c.
b
f â€² (c) = 0
c
2
2. First Derivative Tests â€“ when critical numbers give rise to local minima
and maxima
In order for a critical number x = c to be a local minimum, f (x) must change its monotonicity
at x = c from decreasing to the left of x = c to increasing to the right of x = c. This can be
assured if the sign of f â€² changes from negative to positive.
Theorem 1 (Local minimum – First derivative Test). Let f be defined and differentiable near a
critical number c. A function f has a local minimum at (c, f (c)) if f â€² (x) changes signs at x = c
from negative to positive.
b
f â€² (c) = 0
f â€²â€² (c) > 0
local min
Conversely,
Theorem 2 (Local maximum – First Derivative Test). Let f be defined and differentiable near
a critical number c. A function f has a local maximum at (c, f (c)) if f â€² (x) changes signs at
x = c from positive to negative.
f â€² (c) = 0
f â€²â€² (c) < 0 b local max 3. Example â€“ First derivative test Let f (x) = 2x2 + 2 . x2 + 2 We first determine the critical numbers. Then we check if f â€² changes its sign at the critical numbers. 3 2 +2 y = 2x x2 +2 We calculate that f â€² (x) = How? 2 â€² 2 4x (x2 + 2)2 2 . 2 â€² 2 2 +2)2x +2)(x +2) = 4x(x +2)âˆ’(2x = (x24x . We apply the Product Rule: f â€² (x) = (2x +2) (x +2)âˆ’(2x (x2 +2)2 (x2 +2)2 +2)2 To get critical numbers solve f â€² (x) = 0. Since the denominator is positive we obtain that x = 0 is the only critical number. Note from the formula for f â€² that its sign is detemined by the sign of x. This is because the denominator is always positive. Thus f â€² changes its sign at the critical number x = 0. We conclude that x = 0 is a local minimum. intervals x0 â€² f âˆ’ + monotonicity decreasing â†˜ increasing â†— 4. Second Derivative Test Sometimes at a critical number x = c, it is not diï¬€icult to find the sign of f â€²â€² (c). The negative sign of f â€²â€² (c) indicates that the function f â€² decreases at x = c. Since f â€² (c) = 0 thus f â€² changes its sign from negative to positive at x = c. Conversely, the positive sign of f â€²â€² (c) indicates that f â€² changes its sign at x = c from positive to negative. Thus from the First Derivative Tests we thus obtain: Let f be defined on an interval containing its critical number c. If f â€²â€² (c) > 0, then f has a
local minimum at (c, f (c)). If f â€²â€² (c) < 0, then f has a local maximum at (c, f (c)). 5. Example: second derivative test applied to a quadratic Let )2 ( f (x) = âˆ’ 2x âˆ’ 32 + 85 . We first find the critical numbers. Since f â€² (x) = âˆ’2(2x âˆ’ 23 ), the critical number c = 34 . We now apply the second derivative test. To do this we need to calculate f â€²â€² . 4 f â€²â€² (x) = âˆ’4 , so also f â€²â€² (âˆ’ 34 ) = âˆ’4 < 0 We conclude that at the point ( 34 , f ( 34 )), f reaches its local maximum. f â€² (c) = 0 f â€²â€² (c) < 0 b 6. Concavity and inflections Recall that the concavity of a function f can be determined by the sign of the second derivative f . The graph of y = f (x) is concave up when f â€²â€² (x) > 0 and it is concave down when f â€²â€² (x) < 0. If at x = c the function f (x) changes its concavity, then we have call (c, f (c)) is an inflection point for f (x). Certainly, in this case f â€²â€² (c) = 0. â€²â€² To ensure that a number x = c for which f â€²â€² (c) = 0 gives rise to an inflection, we need to check that f â€²â€² (x) changes its sign at x = c. This can be ensured by checking that f â€²â€² decreases or increases at x = c that is when f â€²â€²â€² (c) Ì¸= 0. Unfortunately, often times it is quite hard to calculate f â€²â€²â€² . (c, f (c)) b f â€²â€² (c) = 0 Example. To the left of x = c the function is convave down as f â€²â€² (x) < 0 for x < c. To the right of x = c, the function f is concave up since f â€²â€² (x) > 0 for x > c. Hence, at x = c the graph
of f (x) has an inflection.
7. Example – Concavities and Inflection Points
Let f (x) = 16 x3 âˆ’ 12 x2 + x âˆ’ 56 . In order to find the inflection points we calculate f â€²â€² (x).
f â€² (x) = ( 16 x3 âˆ’ 21 x2 + x âˆ’ 56 )â€² = 12 x2 âˆ’ x + 1
5
f â€²â€² (x) = x âˆ’ 1.
Since f â€²â€² (x) is negative for x < 1, the graph is concave down. Since f â€²â€² is positive for x > 1,
the graph is concave up when x > 1. Thus f has an inflection point at (1, âˆ’1/6).
b
(1, âˆ’ 16 )
intervals f â€²â€² (x) concavity
x1
âˆ’
+
down
up
8. Example – Inflection Points
Let f (x) = cos(2x) + 13 be defined on the interval (0, Ï€2 ). In order to find the inflection points
we calculate f â€²â€² (x).
f â€² (x) = âˆ’2 sin(2x),
f â€²â€² (x) = âˆ’4 cos(2x).
To find candidates of inflection points we need to solve:
âˆ’4 cos(2x) = 0
restricted to the interval (0, Ï€2 ). We obtain x = Ï€4 that is f â€²â€² ( Ï€4 ) = 0. From the bahavior
of the cosine function, it follows that f â€²â€² = âˆ’4 cos(2x) changes its sign at Ï€4 . More…More
general argument.Alternatively, to show that f â€²â€² changes its sign at Ï€4 it is enough ensure that
f â€²â€² decreases or increases at x = Ï€4 . Since f â€²â€²â€² ( Ï€4 ) = 8 sin(2x)|x= Ï€4 = 8 Ì¸= 0 thus in fact f â€²â€² changes
its sign from negative to positive at x = Ï€4 and thus the function f has an inflection point at
( Ï€4 , 13 ).
6
9. Homework
(1) Give an example of a polynomial P (x) of degree 5 for which x = 1 is a critical number but
not a local minimum, nor a local maximum. (Hint: An example of a polynomial of degree
k is P (x) = (x âˆ’ a)k .
(2) (2â€“5) Determine the intervals on which the function is decreasing and increasing and then
find local minima and maxima. f (x) = (x âˆ’ 2)(x + 3)
(3) f (x) = (x + 1)(x âˆ’ 2)(x + 3)
(4) f (x) = xeâˆ’x
(5) f (x) = xx defined on the interval (0, âˆž).
(6) Let f (x) = cos x + 2 cos2 x defined for Ï€/2 < x < Ï€. Determine the local extrema and inflection points. [Friendly warning: start yesterday...] eed to submit. The graph of y = f â€² (x) is located below the X-axis except for one point where it touches the X-axis at (7, 0). What can you say about the function y = f (x)? Hint. What are the intervals of monotonicity of f ? Does y = f (x) have a crictical point? Does the function whose derivative f â€² (x) is illustrated in the graph below satisfy the above 7 conditions? Purchase answer to see full attachment

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