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Chapter 5
Increasing and Decreasing
Functions
Derivatives are used to describe the shapes of graphs of functions.
5.1 Increasing and Decreasing Functions
For any two values x 1 and x 2 in an interval,
f (x) is increasing iff (x 1 ) < f (x 2 ) if x 1 < x 2 , f (x) is decreasing iff (x 1 ) > f (x 2 ) if x 1.
Derivatives can be used to determine whether a function is increasing, decreasing or constant on an interval:
f (x) is increasing if derivative f ′(x) > 0 , f (x) is decreasing if derivative f ′(x) < 0 , f (x) is constant if derivative f ′(x) = 0.
A critical number, c, is one where f ′(c) = 0 or f ′(c) does not exist; a critical point is (c, f (c)). After locating the critical number(s), choose test values in each interval between these critical numbers, then calculate the derivatives at the test values to decide whether the function is increasing or decreasing in each given interval. (In general, identify values of the function which are discontinuous, so, in addition to critical numbers, also watch for values of the function which are not defined, at vertical asymptotes or singularities (“holes”).)
Exercise 10.1 (Increasing and Decreasing Functions)
- Application: throwing a ball.
162 Chapter 5. Graphs and the Derivative (LECTURE NOTES 10)
P
Q
0
50
1
100
2
150
3
200
4
250
5 time, t (seconds)
(feet)
height, y
(^) R
Figure 5.1 (Graph of function of throwing a ball)
Identify time(s) where height increases and time(s) where height decreases.
(a) Critical time. Change in height stops at c = (i) 0. 75 (ii) 1. 25 (iii) 3. 75 (b) Intervals So there are two intervals to investigate (i) (−∞, 1 .25) (ii) (0, 1 .25) (iii) (1. 25 , 5) (c) Times for increasing and decreasing heights. Height y (i) increases (ii) decreases over (0, 1 .25) interval and (i) increases (ii) decreases over (1. 25 , 5) interval
- Functions.
5
x
y
9/
(e) f(x) = 3x - 2x^3
-15 5 x
y
(a) f(x) = x + 4x - 21
10
2
-5 5 x
y
(b) f(x) = 5x + 4
5
-10 10 x
y
(c) f(x) = 2x + 3x - 36x
100
2
-5 5 x
y
(d) f(x) = 2x - 5x
10
2
decreasing (^) increasing increasing
increasing
increasing
decreasing
164 Chapter 5. Graphs and the Derivative (LECTURE NOTES 10)
since f ′(0) = 6(0 + 3)(0 − 2) is (i) positive (ii) negative function f (x) (i) increases (ii) decreases over (− 3 , −2) interval
since f ′(3) = 6(3 + 3)(3 − 2) is (i) positive (ii) negative function f (x) (i) increases (ii) decreases over (3, ∞) interval
(d) Figure (d): f (x) = 2x^4 − 5 x^2. Since
f ′(x) = 8x^3 + 10x = 8x
( x^2 −
) = 8x
x −
√ 5 4
x +
√ 5 4
(^) = 0,
there are three critical numbers at c = (i) −^54 (ii) −
√ 5 4 (iii)^0 (iv)^
√ 5 4 (v)^
5 4 and so there are four intervals to investigate (i)
( −∞, −
√ (^5) 4
) (ii)
( −
√ (^5) 4 ,^0
) (iii)
( 0 ,
√ (^5) 4
)
(iv)
( 0 ,
√ 5 4
) (v)
(√^5 4 ,^ ∞
)
with four possible test values to check, say: x = (i) − 2 (ii) − 1 (iii) 0 (iv) 1 (v) 2
since f ′(−2) = 8(−2)
( − 2 −
√ 5 4
) ( −2 +
√ 5 4
) (i) positive (ii) negative Notice
√ (^5) 4 ≈^1 .12, so since 8(−2)^ <^ 0,
( − 2 −
√ (^5) 4
) < 0 and
( −2 +
√ (^5) 4
) < 0, f ′(−2) < 0. function f (x) (i) increases (ii) decreases over
( −∞, −
√ 5 4
) interval
since f ′(−1) = 8(−1)
( − 1 −
√ 5 4
) ( −1 +
√ 5 4
) (i) positive (ii) negative Notice
√ (^5) 4 ≈^1 .12, so since 8(−1)^ <^ 0,
( − 1 −
√ (^5) 4
) < 0 and
( −1 +
√ (^5) 4
)
0, f ′(−2) > 0. function f (x) (i) increases (ii) decreases over
( −
√ 5 4 ,^0
) interval
since f ′(1) = 8(1)
( 1 −
√ 5 4
) ( 1 +
√ 5 4
) (i) positive (ii) negative Notice
√ (^5) 4 ≈^1 .12, so since 8(1)^ >^ 0,
( 1 −
√ (^5) 4
) < 0 and
( 1 +
√ (^5) 4
)
0, f ′(−2) < 0. function f (x) (i) increases (ii) decreases over
( 0 ,
√ 5 4
) interval
since f ′(2) = 8(2)
( 2 −
√ 5 4
) ( 2 +
√ 5 4
) (i) positive (ii) negative Notice
√ (^5) 4 ≈^1 .12, so since 8(2)^ >^ 0,
( 2 −
√ (^5) 4
)
0 and
( 2 +
√ (^5) 4
)
0, f ′(−2) > 0. function f (x) (i) increases (ii) decreases over
(√ 5 4 ,^ ∞
) interval
(e) Figure (e): f (x) = 3x^3 − 2 x^4.
Section 1. Increasing and Decreasing Functions (LECTURE NOTES 10) 165
Since f ′(x) = 9x^2 − 8 x^3 = 9x^2
( 1 −
x
) = 0,
there are two critical numbers at c = (i) −^98 (ii) 0 (iii) (^98) and so there are three intervals to investigate (i) (−∞, −3) (ii) (−∞, 0) (iii)
( 0 , (^98)
) (iv)
( 9 8 ,^ ∞
) (v) (0, ∞) with three possible test values to check, say: x = (i) − 1 (ii) 0 (iii) 1 (iv) 98 (v) 2
since f ′(−1) = 9(−1)^2
( 1 − 89 (−1)
) is (i) positive (ii) negative function f (x) (i) increases (ii) decreases over (−∞, 0) interval
since f ′(1) = 9(1)^2
( 1 − 89 (1)
) is (i) positive (ii) negative function f (x) (i) increases (ii) decreases over
( 0 , (^98)
) interval
since f ′(2) = 9(2)^2
( 1 − 89 (2)
) is (i) positive (ii) negative function f (x) (i) increases (ii) decreases over
( (^9) 8 ,^ ∞
) interval
- Functions whose derivatives do not exist at some values.
5 x
y f(x) = x
critical point
1/
(a) function value f(0) defined, but derivative value f’(0) does not exist so a critical number at 0
5
x
y
asymptote
(b) function value f(0) undefined, so derivative f’(0) does not exist so NOT a critical number at 0
f(x) = x-1/
Figure 5.3 (Functions whose derivatives do not exist at some values)
(Type all functions into your calculator using Y= and set WINDOW -5 5 1 -5 5 1 1.
(a) Figure (a): f (x) = x (^13) . Since f ′(x) =
x−^
2 (^3) =
x)^2
as x → 0, slope of tangent, f ′, becomes vertical (does not exist), limx→ 0 f ′(x) = (i) −∞ (ii) 0 (iii) ∞
Section 2. Relative Extrema (LECTURE NOTES 10) 167
If function f has a relative extremum at c, then c is either a critical number or an endpoint. First derivative test for locating relative extrema in (a, b):
f (c) is relative maximum if f ′(x) positive in (a, c), negative in (c, b) f (c) is relative minimum if f ′(x) negative in (a, c), positive in (c, b)
Although treated in a similar manner, allowances are made for identifying relative extrema for functions with discontinuities.
Exercise 10.2 (Relative Extrema)
- Critical points, endpoints and relative extrema.
0
2
4
6
1
8
3
10
5
y
-5 x
A
B
C
D
E
F
G
H
vertical tangent line at point C
Figure 5.4 (Critical points, endpoints and extrema)
(a) Point A where x = −5 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum because, as suggested by the text, function heads down after point A; however, some would dispute this because if the function was previously heading down to A, then A would neither be a maximum nor minimum; in other words, it really is not clear what A is because there are no points on “both sides” of A to be able to “really” decide if it is a maximum or minimum (b) Point B where x = − 3 .5 is (i) a critical point (ii) an endpoint
168 Chapter 5. Graphs and the Derivative (LECTURE NOTES 10)
(iii) neither a critical point nor endpoint because tangent line at B is zero, f ′(B) = 0 which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum (c) Point C where x = − 2 .7 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint because although f (C) is defined, f ′(C) does not exist which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum because slope (derivative) f ′(x) is positive both before and after critical point C (d) Point D where x = −2 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint because tangent line at D is zero, f ′(D) = 0 which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum (e) Point E where x = 0.5 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint because although f (C) is defined, f ′(C) does not exist which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum (f) Point F where x = 3 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint because tangent line at D is zero, f ′(D) = 0 which is (i) a relative minimum
170 Chapter 5. Graphs and the Derivative (LECTURE NOTES 10)
(ii) an endpoint (iii) neither a critical point nor endpoint which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum because function heads down after point A (b) Point B where x = −4 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint because tangent line at B is zero, f ′(B) = 0 which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum because function decreases down to B then increases afterwards (c) Point C where x = − 1 .5 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint because tangent line at C is zero, f ′(C) = 0 which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum because function increases up to C then decreases afterwards (d) Point D where x = −1 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint because limit at x = −1 does not exist since limx→− 1 − f (x) = 6. 8 6 = limx→− 1 + f (x) = 5 which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum because function decreases down to D then increases afterwards (e) Point E where x = 0 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint
Section 2. Relative Extrema (LECTURE NOTES 10) 171
because tangent line at E is zero, f ′(E) = 0 which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum because function increases up to C then decreases afterwards (f) Point F where x = 3 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint because function approaches infinity which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum because function approaches infinity, not any particular large number (g) Point G where x = 3.4 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint because tangent line at G is zero, f ′(G) = 0 (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum because function decreases down to B then increases afterwards (h) Point H where x = 4.2 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint which is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum because function increases both before and after H (i) Point I where x = 5 is (i) a critical point (ii) an endpoint (iii) neither a critical point nor endpoint because the function is defined on an interval which is open at this end which is (i) a relative minimum
Section 2. Relative Extrema (LECTURE NOTES 10) 173
(b) First derivative test. At critical number c = −2, sign of derivative f ′(x) goes from (i) negative to positive (ii) positive to negative (iii) negative to negative (iv) positive to positive
and so, according to first derivative test, there is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum at critical number c = −2, and since f (−2) = (−2)^2 + 4(−2) − 21 = −25, at critical point (c, f (c)) = (− 2 , −25).
- First derivative test: f (x) = 2x^3 + 3x^2 − 36 x revisited
zero slope
negative slope
-10^10 x
(^100) y
2
zero slope
positive slope
positive slope
test point critical point
relative minimum
relative maximum
Figure 5.7 (First derivative test: f (x) = 2x^3 + 3x^2 − 36 x)
GRAPH using Y 3 = 2x^3 + 3x^2 − 36 x, with WINDOW -10 10 1 -100 100 1 1
(a) Critical numbers and intervals. Recall, since
f ′(x) = 6x^2 + 6x − 36 = 6(x^2 + x − 6) = 6(x + 3)(x − 2) = 0,
there are two critical numbers at c = (i) − 3 (ii) 2 (iii) 6 and so there are three intervals to investigate (i) (−∞, −3) (ii) (−∞, 2) (iii) (− 2 , 3) (iv) (− 3 , 2) (v) (2, ∞) with three possible test values to check, say: x = (i) − 4 (ii) − 3 (iii) 0 (iv) 2 (v) 3
174 Chapter 5. Graphs and the Derivative (LECTURE NOTES 10)
since f ′(−4) = 6(−4 + 3)(− 4 − 2) is (i) positive (ii) negative function f (x) (i) increases (ii) decreases over (−∞, −3) interval
since f ′(0) = 6(0 + 3)(0 − 2) is (i) positive (ii) negative function f (x) (i) increases (ii) decreases over (− 3 , −2) interval
since f ′(3) = 6(3 + 3)(3 − 2) is (i) positive (ii) negative function f (x) (i) increases (ii) decreases over (3, ∞) interval so summarizing: interval (−∞, −3) (− 3 , 2) (2, ∞) test value x = − 5 x = 0 x = 3 f ′(x) = 6x^2 + 6x − 36 f ′(−5) = 84 f ′(0) = − 36 f ′(3) = 36 sign of f ′(x) positive negative positive
(b) First derivative test. At critical number c = −3, sign of derivative f ′(x) goes from (i) negative to positive (ii) positive to negative (iii) negative to negative (iv) positive to positive and so, according to first derivative test, there is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum at critical number c = −3, and since f (−3) = 2(−3)^3 + 3(−3)^2 − 36(−3) = 81, at critical point (c, f (c)) = (− 3 , 81).
At critical number c = 2, sign of derivative f ′(x) goes from (i) negative to positive (ii) positive to negative (iii) negative to negative (iv) positive to positive and so, according to first derivative test, there is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum at critical number c = 2, and since f (2) = 2(2)^3 + 3(2)^2 − 36(2) = −44, at critical point (c, f (c)) = (2, −44).
- First derivative test: f (x) = 3x^3 − 2 x^4 revisited
176 Chapter 5. Graphs and the Derivative (LECTURE NOTES 10)
(ii) positive to negative (iii) negative to negative (iv) positive to positive and so, according to first derivative test, there is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum at critical number c = 0, and since f (0) = 3(0)^3 − 2(0)^4 = 0, at critical point (c, f (c)) = (0, 0).
At critical number c = 98 , sign of derivative f ′(x) goes from (i) negative to positive (ii) positive to negative (iii) negative to negative (iv) positive to positive and so, according to first derivative test, there is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum at critical number c = 98 , and since f
( 9 8
) = 3
( 9 8
) 3 − 2
( 9 8
) 4 = 21872048 , at critical point (c, f (c)) =
( 9 8 ,^
2187 2048
) .
- First derivative test: f (x) = x^3 (5.5)x^ revisited Type Y=, then Y 1 = x^3 (5.5)x^ , WINDOW -2.5 1 1 -0.5 0.5 1 1, then GRAPH
(a) Derivative.
Let u(x) = x^3 and v(x) = 5. 5 x then u′(x) = (i) 3 x^2 (ii) 3 x (iii) e^2 x and v′(x) = (i) 5. 5 x^ (ii) (ln 5.5) (iii) (ln 5.5) 5. 5 x
and so v(x)u′(x) = (i) 5. 5 x (ii) 3 x^3 (iii) (5. 5 x) (3x^2 )
and u(x)v′(x) = (i) (x^3 ) (ln 5.5) 5. 5 x (ii) (ln 5.5) 5. 5 x (iii) (3x^3 ) (ln 5.5)
Section 2. Relative Extrema (LECTURE NOTES 10) 177
and so f ′(x) = v(x) · u′(x) + u(x) · v′(x) = (i) (5. 5 x) (3x^2 ) + (3x^3 ) (ln 5.5) (ii) (5. 5 x) + (3x^3 ) (ln 5.5) 5. 5 x (iii) (5. 5 x) (3x^2 ) + (x^3 ) (ln 5.5) 5. 5 x which simplifies to
f ′(x) =
( x^2
) (5. 5 x) (3 + x ln 5.5)
Type Y=, then Y 2 =
( x^2
) (5. 5 x^ ) (3 + x ln 5.5), WINDOW -2.5 1 1 -0.5 0.5 1 1, then GRAPH (b) Critical numbers and intervals. Since f ′(x) =
( x^2
) (5. 5 x) (3 + x ln 5.5) = 0, there are two critical numbers at c = (i) − (^) ln 5^3. 5 ≈ − 1. 76 (ii) 0 (iii) (^) ln 5^3. 5 (Since 3 + x ln 5.5 = 0, x = − (^) ln 5^3. 5 ≈ − 1 .76) and so there are three intervals to investigate (i)
( −∞, − (^) ln 5^3. 5
) (ii)
( − (^) ln 5^3. 5 , 0
) (iii) (0, 3) (iv) (3, 5 .5) (v) (0, ∞) with three possible test values to check, say: x = (i) − 2 (ii) − (^) ln 5^3. 5 (iii) − 1 (iv) 0 (v) 1
since f ′(−2) = ((−1)^2 ) (5. 5 −^2 ) (3 − 2 ln 5.5) is (i) positive (ii) negative VARS Y-VARS ENTER Y 2 ENTER (−2) ENTER gives approximately -0.054, which is negative function f (x) (i) increases (ii) decreases over
( −∞, − (^) ln 5^3. 5
) interval
since f ′(−1) = ((−1)^2 ) (5. 5 −^1 ) (3 − ln 5.5) is (i) positive (ii) negative VARS Y-VARS ENTER Y 2 ENTER (−1) ENTER gives approximately 0.236, which is positive function f (x) (i) increases (ii) decreases over
( − (^) ln 5^3. 5 , 0
) interval
since f ′(1) = ((1)^2 ) (5. 51 ) (3 + ln 5.5) is (i) positive (ii) negative VARS Y-VARS ENTER Y 2 ENTER (1) ENTER gives approximately 25.876, which is positive function f (x) (i) increases (ii) decreases over (0, ∞) interval so summarizing:
interval
( −∞, − (^) ln 5^3. 5
) ( − (^) ln 5^3. 5 , 0
) (0, ∞) test value x = − 2 x = − 1 x = 1 f ′(x) = (x^2 ) (5. 5 x) (3 + x ln 5.5) f ′(−2) ≈ − 0. 054 f ′(−1) ≈ 0. 236 f ′(1) ≈ 25. 877 sign of f ′(x) negative positive positive
Section 2. Relative Extrema (LECTURE NOTES 10) 179
(c) At critical number c = 5, there is a (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum
- Application: throwing a ball. A ball is thrown upwards with an initial velocity of 32 feet per second and from an initial height of 150 feet. A function relating height, f (t), to time, f , when throwing this ball is:
f (t) = − 12 t^2 + 32t + 150
Find maximum height, f (t), and time, t, ball reaches maximum height. GRAPH using Y 5 = − 12 x^2 + 32x + 150, with WINDOW 0 6 1 0 250 1 1
(a) Derivative f ′(t) = −12(2)t^2 −^1 + 32(1)t^1 −^1 = 0 = (i) − 24 t + 32 (ii) 24 t − 32 (iii) 24 t^2 + 32 (b) Critical numbers and intervals. Since f ′(x) = − 24 t + 32 = 0, there is a critical number at c = − (^) −^3224 = (i) 43 (ii) −^43 (iii) (^34) and so there are two intervals to investigate (i)
( −∞, (^43)
) (ii)
( (^4) 3 ,^ ∞
) (iii) (− 2 , ∞) with two possible test values (in each interval) to check, say: x = (i) 0 (ii) 1 (iii) 2
and since f ′(0) = −24(0) + 32 = 32 is positive, function f (x) (i) increases (ii) decreases over
( −∞, (^43)
) interval
and f ′(2) = −24(2) + 32 = −16 is negative, function f (x) (i) increases (ii) decreases over
( 4 3 ,^ ∞
) interval so summarizing:
interval
( −∞, (^43)
) ( 4 3 ,^ ∞
)
test value x = 0 x = 2 f ′(t) = − 24 t + 32 f ′(0) = 32 f ′(2) = − 16 sign of f ′(x) positive negative
(c) First derivative test. At critical number c = 43 , sign of derivative f ′(t) goes from (i) negative to positive
180 Chapter 5. Graphs and the Derivative (LECTURE NOTES 10)
(ii) positive to negative (iii) negative to negative (iv) positive to positive
and so, according to first derivative test, there is (i) a relative minimum (ii) a relative maximum (iii) not a relative extremum at critical number c = 43 , and since f
( 4 3
) = − 12
( 4 3
) 2
( 4 3
)
- 150 = 5143 , at critical point (c, f (c)) =
( 4 3 ,^
514 3
) . (d) Results In other words, ball reaches maximum height of (i) 43 (ii) 5143 (iii) (^5143) at time (i) 43 (ii) −^43 (iii) (^34)
