Section 3.4
7) f(x) = x4 – 24x2 + 40
7a) Domain of all polynomials is all real numbers. There is no Algebra that is needed to find the domain
for this problem. If you try some Algebra you will likely find the x-intercept and not a number that is
part of the domain.
Answer: (−∞, ∞)
7b) x-intercept
The Algebra for this is a bit messy. You would need u-substitution followed by the quadratic formula.
You may use your calculator to find these x-intercepts.
Answer: (-4.71, 0) (-1.34, 0) (1.34, 0) (4.71, 0) (Used calculator)
7c) f(0) = (0)4 – 24(0)2 + 40
= 0 – 0 + 40
= 40
Answer: y-intercept (0,40)
7d) f’(x) = 4x3 – 48x
4x3 – 48x = 0
4x(x2 – 12) = 0
4x = 0
x2 – 12 = 0
x = 0/4
x1 = 12
x=0
𝑥 = ±√12 = ±√4 ∗ √3
𝑥 = ±2√3
(−∞, −2√3)
(−2√3, 0)
(0,2√3)
(2√3, ∞)
Test x = -4
test x = -1
test x = 1
test x = 4
f’(-4) = -64
decreasing
f’(-1) = 44
increasing
f’(1) = -44
decreasing
f’(4) = 64
increasing
Answer: increasing (−2√3, 0) ∪ (2√3, ∞)
7e) decreasing (−∞, −2√3) ∪
(0,2√3)
7f) Have maximum at x = 0 and a minimum of x = -2√3 𝑎𝑛𝑑 2√3
I used the value feature to get the y-values. That is I graphed and then chose 2nd calc – value and found
𝑓(−2√3) = −104
min
f(0) = 40
max
𝑓(2√3)= -104
min
Answer: rel max y = 40 when x = 0. rel min y = -104 when x = 2√3 and −2√3
7g)
f(x) = x4 – 24x2 + 40
f’(x) = 4x3 - 48x
f”(x) = 12x2 – 48
12x2 – 48 = 0
12(x2 – 4 ) = 0
12(x+2)(x-2) = 0
12 = 0
x+2=0
x–2=0
No sol
x = -2
x=2
(−∞, −2)
(-2,2)
(2, ∞)
Test x = -3
Test x = 0
test x = 3
f”(-3) = 60
f”(0) = -48
f”(3) = 60
concave down
concave up
concave down
7g) Answer: concave up (−∞, −2) ∪ (2, ∞)
7h) Answer: concave down (-2,2)
7i) x-coordinate of inflection points x = 2 and x = -2.
Find each y using value feature of calculator when the original problem is in y1 graph.
f(2) = -40
f(-2) = -40
Answer: inflection points (2,-40) and (-2,-40)
7j) I used a program to graph the function. I marked the points with dots, but didn’t put numbers next
to the points as some of the numbers are decimals. Let me know if you can’t tell what the values of the
points are.
9) f(x) = x5 – 15x3
9a) Domain of all polynomials is all real numbers. There is no Algebra that is needed to find the domain
for this problem. If you try some Algebra you will likely find the x-intercept and not a number that is
part of the domain.
Answer: (−∞, ∞)
9b) x5 – 15x3 = 0
x3(x2 – 15) = 0
x3 = 0
x2 – 15 = 0
x=0
x2 = 15
x = ±√15
Answer: x-intercepts (0,0) ∪ (−√15, 0) ∪ (√15, 0)
9c) y-intercept f(0) = (0)5 – 15(0)2 = 0
Answer = y-intercept (0,0)
9d) f’(x) = 5x4 – 45x2
5x4 – 45x2 = 0
5x2(x2 – 9) = 0
5x2(x+3)(x-3) = 0
5x = 0
x+3 = 0
x–3=0
x=0
x = -3
x=3
(−∞, −3)
(-3, 0)
(0,3)
(3, ∞)
Test x = -4
test x = -2
test x = 2
test x = 4
f’(-4) = 560
f’(-2) = -100
f’(1) = -100
f’(4) = 560
increasing
decreasing
decreasing
increasing
9d) Answer: Inc (−∞, −3) ∪ (3, ∞)
9e) Answer: dec (−3,0) ∪ (0,3)
9f)
x coord of rel max x = -3 find
Find y-coord using value feature on graph f(-3) = 162
x coord of rel min at x = 3
Find y-coord using value feature on graph f(3) = -162
9f) Answer: rel max of y = 162 at x = -3 rel min of y = -162 when x = 3
f(x) = x5 – 15x3
9g)
f’(x) = 5x4 – 45x2
f”(x) = 20x3 – 90x
20x3 – 90x = 0
10x( 2x2 – 9) = 0
10x = 0
2x2 – 9 = 0
x=0
2x2 = 9
x2 = 9/2
(−∞,
𝑥=±
√9
√2
𝑥=±
3
√2
∗
√2 √2
= ±
−3√2
)
2
3
√2
=±
3√2
2
−3√2
, 0)
2
3√2
)
2
3√2
, ∞)
2
(
(0,
test x = --5
test x = -1
test x = 1
test x = 5
f’’(-5) = -2050
f’’(-1) = 70
f’’(1) = -70
f’’(5) = 2050
concave down
concave up
concave down
concave up
9g) Answer: concave up (−
3√2
3√2
, 0) ∪ ( 2 , ∞)
2
9h) Answer: concave down (−∞,
−3√2
)∪
2
(0,
3√2
)
2
(
9i) x-coordinates of inflection points x = 0
x=
−3√2
2
𝑥=
3√2
2
Computer y-coordinates (use value feature on calculator a decimal will be okay as this is quite messy to
do by hand)
x=0
x=
f(0) = 0
f(
−3√2
2
−3√2
)
2
= 100.23
𝑥=
3√2
2
3√2
)
2
= -100.23
f(
−3√2
3√2
, 100.23) ( 2 , −100.23)
2
Answer: inflection points (0,0) (
9f) I used a program to graph the function. I marked the points with dots, but didn’t put numbers next
to the points as some of the numbers are decimals. Let me know if you can’t tell what the values of the
points are.
11) f(x) = x3 – 27x
11a) Domain of all polynomials is all real numbers. There is no Algebra that is needed to find the
domain for this problem. If you try some Algebra you will likely find the x-intercept and not a number
that is part of the domain.
Answer: (−∞, ∞)
11b) x3 – 27x = 0
x(x2 – 27) = 0
x=0
𝑥 2 − 27 = 0
x=0
𝑥 2 = 27
𝑥 = ±√27 = ±√9√3 = ±3√3
11b) Answer (0,0) ∪ (−3√3, 0) ∪ (3√3, 0)
11c) f(0) = (0)3 – 27(0) = 0
11c) Answer: (0,0)
11d)
f’(x) = 3x2 – 27
3x2 – 27 = 0
3(x2 – 9) = 0
3(x+3)(x-3) = 0
3=0
x+3=0
x–3=0
No solution
x = -3
x=3
(−∞, −3)
(−3,3)
(3, ∞)
Test x = -4
test x = 0
test x = 4
f’(-4) = 21
f’(0) = -27
f’(4) = 21
increasing
decreasing
increasing
11d) Answer increasing (−∞, −3) ∪ (3, ∞)
11e) Answer decreasing (-3,3)
11f) x-coord of rel max x = -3
Computation of y using value feature on calculator f(-3) = 54
x-coord of rel min x = 3
Computation of y using value feature on calculator f(3) = -54
11f) Answer: rel max of y = -54 when x = 3 rel min of y = 54 when x = -3
11g) f(x) = x3 - 27x
f’(x) = 3x2 – 27
f’’(x) = 6x
6x = 0
x=0
(−∞, 0)
(0, ∞)
Test x = -1
test x = 1
f’’(-1) = -6
f’’(1) = 6
concave down
concave up
11g) answer: concave up (0, ∞)
11h) answer (−∞, 0)
11i) x-coord of inflection point x = 0
Compute y-coord using value feature on calculator f(0) = 0
11i) Answer inflection point (0,0)
11j) I used a program to graph the function. I marked the points with dots, but didn’t put numbers next
to the points as some of the numbers are decimals. Let me know if you can’t tell what the values of the
points are.
15j) the graph changes concavity at x=3, however the function is not defined at x=3. When the function
is not defined for the value of x where there should be an inflection point there is no inflection pont.
15j) answer none
15k)
2
17) 𝑓(𝑥) = 𝑥 2 +6𝑥−7
x2 + 6x – 7 = 0
17a)
(x+7)(x – 1) = 0
x+7=0
x–1=0
x = -7
x = 1 (these are x-coord of the vertical asymptotes)
plot on a number line along with the infinities
−∞
-7
∞
1
This creates the intervals
(−∞, −7)
(-7,1)
(1,∞)
17a) Answer: 𝑑𝑜𝑚𝑎𝑖𝑛 (−∞, −7) ∪ (−7,1) ∪ (1, ∞) vertical asymptote x = -7 and x = 1
17b) to find the x-intercept of a fraction just set the numerator equal to 0
2=0
There is no solution as this is a false statement
17b) Answer: no x-intercept
17c)
y-intercept 𝑓(0) =
2
02 +6(0)−7
17c) Answer y-intercept (0,
17d) horizontal asym lim
=
2
−7
−2
)
7
2
𝑥→∞ 𝑥 2 +6𝑥−7
1
= lim
1
2∗ 2
𝑥
1
1 = lim
𝑥→∞ 𝑥 2 ∗ 2 +6𝑥∗ 2 −7∗ 2
𝑥
𝑥
𝑥
𝑥→∞
2
𝑥2
6 7
1+ − 2
𝑥 𝑥
= lim
𝑥→∞
17d) Answer: horizontal asymptote y = 0
2
∞2
6
7
1+ − 2
∞ ∞
0
= 1+0−0 = 0
17e) 𝑓 ′ (𝑥) =
(𝑥 2 +6𝑥−7)(0)−2(2𝑥+6) −2(2𝑥+6)
=(𝑥 2 +6𝑥−7)2
(𝑥 2 +6𝑥−7)2
(𝑥 2 + 6𝑥 − 7)2 = 0
-2(2x+6) = 0
-2 = 0
2x+6=0
x2 + 6x – 7 = 0
No sol
2x = -6
(x+ 7)(x-1) = 0
x = -3
x = -7
x=1
(−∞, −7)
(-7,-3)
(-3,1)
(1, ∞)
test x = -8
test x = -4
test x = 0
test x = 2
f’(-8) = .25
f’(-4) = .02
f’(0) = -.24
f’(2) = -.25
increasing
increasing
decreasing
decreasing
17e) Answer: increasing (−∞, −7) ∪ (−7, −3)
17f) Answer: decreasing (−3,1) ∪ (1, ∞)
17g) x-coord of max x = -3
2
2
Computation of y f(-3) = (−3)2 +6(−3)−7 = −16 =
−1
8
There is no relative min
17g) Answer: rel max of y =-1/8 when x = -3 rel min none
17h) 17i) 17j) skip for problem 17
17k)
1
19) 𝑓(𝑥) = 𝑥 2 −4
19a) x2 – 4 = 0
(x +2)(x-2) = 0
x = -2,2 plot on number line with infinity to get intervals needed for domain.
−∞
-2
(−∞, −2)
∞
2
(-2,2)
(2,∞)
19a) Answer: Domain (−∞, −2) ∪ (−2,2) ∪ (2, ∞) Vertical asymptote x = 2 and x = -2
19b) to find the x-intercept of a fraction just set the numerator equal to 0
2=0
There is no solution as this is a false statement
19b) Answer: no x-intercept
1
1
19c) f(0) = 02 −4 = −4 =
−1
4
19c) answer y-intercept (0,
−1
)
4
19d) horizontal asymptote
1
2 −4
𝑥
𝑥→∞
lim
lim
𝑥→∞
1
∞2
4
1− 2
∞
1
=
=
1∗ 2
𝑥
lim
1
1
𝑥→∞ 𝑥 2 ∗ 2 −4∗ 2
𝑥
𝑥
0
1−0
0
=1=0
19d) Answer y = 0
= lim
𝑥→∞
1
𝑥2
4
1− 2
𝑥
=
19e) 𝑓 ′ (𝑥) =
(𝑥 2 −4)0−2𝑥(1)
(𝑥 2 −4)2
−2𝑥
= (𝑥 2 −4)2
-2x = 0
(x2 – 4)2 = 0
x = 0/-2
x2 – 4 = 0
x=0
(x+2)(x-2) = 0
x = -2 x = 2
(−∞, −2)
(-2,0)
(0,2)
(2,∞)
Check x = -3
check x = -1
check x = 1
check x = 3
f’(-3) = .24
f’(-1) = .22
f’(1) = -.22
f’(3) = -.24
increasing
increasing
decreasing
decreasing
19e) Answer: Increasing (−∞, −2) ∪ (−2,0)
19f) Answer: decreasing (0,2) ∪ (2, ∞)
19g) x-coord of relative max x = 0
Compute y
f(0) =
1
02 −4
=
1
−4
=
−1
4
There is no rel min
19g) Answer rel max of y = -1/4 when x = 0 there is no rel min
19h) 19i) 19j) skip for problem 19
19k)
21) f(x) = 𝑒 𝑥
2
a) Find the domain
There won’t be any algebra that we can do to find the domain of any of the e problems. The domain of
each “e” problem will be (−∞, ∞)
21a) Answer: (−∞, ∞)
b) Find the x-intercept(s), if any
2
𝑒 𝑥 = 0 this has no solution, so there is no x-intercept
21b) answer no x-intercept
2
21c) f(0) = 𝑒 0 = 1
21c) Answer y-intercept (0,1)
21d) Find all vertical asymptotes and horizontal asymptotes
We will find these by inspecting the graph we get on our calculator. This graph doesn’t have either a
vertical nor a horizontal asymptote.
21d) Answer none
21e) Find the interval(s) where the graph of the function is increasing
f’(x) = 2x𝑒 𝑥
2
2
2x = 0
𝑒𝑥 = 0
x=0
no solution
−∞
∞
0
(−∞, 0)
(0, ∞)
Test x = -1
test x = 1
f’(-1) – 5.44
f’(1) = 5.44
decreasing
increasing
21e) Answer increasing (0, ∞)
21f) Answer decreasing (−∞, 0)
2g) Find all relative maxima and relative minima
x-coord of rel min x = 0
compute y, may use value feature from calculator f(0) = 1
There is no rel maz
21g) answer rel max – none
rel min of y = 1 when x = 0
21h) Find the interval(s) where the graph of the function is concave up (if any)
f’(x) = 2x𝑒 𝑥
2
2
f’’(x) = 2𝑒 𝑥 + 2𝑥 ∗ 2𝑥𝑒 𝑥
2
2
f’’(x) = 2𝑒 𝑥 (1 + 2𝑥 2 )
2
2𝑒 𝑥 (1 + 2𝑥 2 ) = 0
2
2𝑒 𝑥 = 0
1 + 2x2 = 0
No solution
2x2 = -1
x2 = -1/2
−1
x = ±√ 2
This gives an i, so I can ignore the answer
There are no numbers other than the infinities to graph on my number line. There will be only one
interval.
−∞
∞
(−∞, ∞)
Test x = 0
2
f’’(0) = 2𝑒 0 (1 + 2(0)2 ) = 2
graph is always concave up and never concave down
21h) answer concave up (−∞, ∞)
21i) Find the interval(s) where the graph of the function is concave down (if any)
21i) answer: never concave down
j) Find all inflection points (if any)
21j) since the graph does not change concavity there are no inflection points.
k) Sketch a graph
23) f(x) = xex
23a) Find the domain
There won’t be any algebra that we can do to find the domain of any of the e problems. The domain of
each “e” problem will be (−∞, ∞)
23a Answer: (−∞, ∞)
23b) Find the x-intercept(s), if any
xex = 0
x=0
ex = 0 (no solution)
23b) answer (0,0)
23c) Find the y-intercept, in there is one
f(0) = 0*e0 = 0
23c) Answer y-intercept (0,0)
23d) Find all vertical asymptotes and horizontal asymptotes
We will find these by inspecting the graph we get on our calculator. This graph doesn’t have either a
vertical asymptote. However the left side of the graph flattens out around the x-axis. It would be
correct to say that there is a horizontal asymptote at the x-axis which has the equation y=0.
23d) Answer – no vertical asymptote
Horizontal asymptote y = 0
23e) Find the interval(s) where the graph of the function is increasing
f’(x) = xex + 1ex
xex + 1ex = 0
ex(x + 1) = 0
ex = 0
x+1=0
no solution
x = -1
(−∞. −1)
(−1, ∞)
Check x = -2
check x = 0
f’(-2) = -.14
f’(0) = 1
decreasing
increasing
23e) answer increasing (−1, ∞)
23f) Find the interval(s) where the graph of the function is decreasing
23f) Answer decreasing (−∞. −1)
23g) Find all relative maxima and relative minima
x-coord of rel min x = -1
computation of y coord f(-1) = -1e-1 = -1/e
there is no relative max
23g) answer rel max none
rel min (-1, -1/e)
23h) Find the interval(s) where the graph of the function is concave up (if any)
f’(x) = xex + 1ex
f’’(x) = xex + 1ex + 1ex
f’’(x) = ex(x + 2)
ex(x+2) = 0
ex = 0
x+2 = 0
no solution
x = -2
(−∞, −2)
test x = -3
(−2, ∞)
test x = 0
f’’(-3) = -.05
f’’(0) = 2
concave down
concave up
23h) answer concave up (−2, ∞)
23i) Find the interval(s) where the graph of the function is concave down (if any)
23i) answer concave down (−∞, −2)
23j) Find all inflection points (if any)
x-coord of inflection point x = -2
computation of y f(-2) = -2e-2 = -2/e2
−2
23j) answer inflection point (−2, 𝑒 2 )
23k) Sketch a graph
25) f(x) = x2ex
25a) Find the domain
There won’t be any algebra that we can do to find the domain of any of the e problems. The domain of
each “e” problem will be (−∞, ∞)
25a)Answer: (−∞, ∞)
25b) Find the x-intercept(s), if any
x2ex = 0
x2 = 0
ex = 0
x=0
no solution
25b) answer x-intercept (0,0)
25c) Find the y-intercept, in there is one
f(0)= 02e0 = 0
25c) answer y-intercept (0,0)
25d) Find all vertical asymptotes and horizontal asymptotes
We will find these by inspecting the graph we get on our calculator. This graph doesn’t have either a
vertical asymptote. However the left side of the graph flattens out around the x-axis. It would be
correct to say that there is a horizontal asymptote at the x-axis which has the equation y=0.
25d) Answer – no vertical asymptote
Horizontal asymptote y = 0
25e) Find the interval(s) where the graph of the function is increasing
f’(x) = 2xex + x2ex
2xex + x2ex = 0
xex(2 + x ) = 0
x=0
ex = 0
2+x=0
no solution
x = -2
(−∞, −2)
(-2, 0)
(0, ∞)
Test x = -3
test x = -1
test x = 1
f’(-3) = .15
f’(-1) = -.37
f’(1) = 8.15
increasing
decreasing
increasing
25e) answer increasing (−∞, −2) ∪ (0, ∞)
25f) Find the interval(s) where the graph of the function is decreasing
25f) answer decreasing (-2,0)
25g) Find all relative maxima and relative minima
x coordinate of rel max x = 0
computation of y f(0) = 0 (used value feature on calculator)
x coordinate of rel min x = -2
computation of y f(-2) = (-2)2e-2 = 4/e2
25g) answer rel max of y = 0 when x = 0 rel min of y = 4/e2 when x = -2
25h) Find the interval(s) where the graph of the function is concave up (if any)
f’’(x) = 2ex + 2xex + x2ex + 2xex
f’’(x) = ex(2 + 2x + x2 + 2x)
f’’(x) = ex(x2 + 4x + 2)
ex(x2 + 4x + 2) = 0
ex = 0
x2 + 4x + 2 = 0
no solution
use quad formula x =
−4±√4 2 −4∗1∗2
2∗1
=
−4±√8 −4±2√2
= 2 =
2
−∞
-2 - √2
(−∞, −2 − √2)
(-2 - √2 , -2 + √2)
(-2 + √2, ∞)
Test x = -4
test x = -1
test x = 0
f’’(-4) = -.03
f’’(-1) = -.37
f’’(0) = 2
concave up
concave down
concave up
-2 ±√2
∞
-2 + √2
25h) answer concave up (−∞, −2 − √2) ∪ (-2 + √2, ∞)
25)i) Find the interval(s) where the graph of the function is concave down (if any)
25i) answer concave down (-2 - √2, -2 + √2)
25j) Find all inflection points (if any)
x coord of inflection points x = 2 - √2
x = -2 + √2
I will use the value feature and give decimal y coords
25j) answer inflection points (2 - √2, .38 ) ∪(-2 + √2, .19 )
25k) Sketch a graph
27) f(x) = 2xe-3x
a) Find the domain
There won’t be any algebra that we can do to find the domain of any of the e problems. The domain of
each “e” problem will be (−∞, ∞)
Answer: (−∞, ∞)
27b) Find the x-intercept(s), if any
2xe-3x = 0
2x = 0
e-3x = 0
X=0
no solution
27b) answer x-intercepts (0,0)
27c) Find the y-intercept, in there is one
f(0) = 2(0)e-3*0 = 0
27c) answer y-intercept (0,0)
27d) Find all vertical asymptotes and horizontal asymptotes
We will find these by inspecting the graph we get on our calculator. This graph doesn’t have either a
vertical asymptote. However the left side of the graph flattens out around the x-axis. It would be
correct to say that there is a horizontal asymptote at the x-axis which has the equation y=0.
27d) Answer – no vertical asymptote
Horizontal asymptote y = 0
27e) Find the interval(s) where the graph of the function is increasing
f’(x) = 2e-3x + -3*2xe-3x
f’(x) = 2e-3x(1 -3x)
2e-3x = 0
1 – 3x = 0
No solution
1 = 3x
x = 1/3
1
1
(−∞, 3)
(3 , ∞)
Check x = 0
check x = 1
f’(0) = 2
f’(1) = -.19
increasing
decreasing
1
3
27e) answer increasing (−∞, )
27f) Find the interval(s) where the graph of the function is decreasing
1
27f) answer decreasing (3 , ∞)
27g) Find all relative maxima and relative minima
x coord of rel max x = 1/3
computation of y
1
𝑓( )
3
=
1
1
2 ( ) 𝑒 −3∗3
3
f
=
2 −1
𝑒
3
=
2
3𝑒
there is no rel min
2
27g) answer rel max of y = (3𝑒) when x = 1/3
rel min none
27h) Find the interval(s) where the graph of the function is concave up (if any)
f’’(x) = -3*2e-3x + -6e-3x + (-3)(-6x)e-3x
f’’(x) = -6e-3x – 6e-3x + 18xe-3x
f’’(x) = -12e-3x + 18xe-3x
f’’(x) = -6e-3x(2-3x)
-6e-3x = 0
2 – 3x = 0
No solution
2 = 3x
x = 2/3
2
2
(−∞, 3)
(3 , ∞)
Check x = 0
check x = 1
f’’(0) = -12
f’’(1) = .30
2
27h) answer concave up (3 , ∞)
27)i) Find the interval(s) where the graph of the function is concave down (if any)
2
27i) answer concave down (−∞, 3)
27j) Find all inflection points (if any)
2
2
2
4
4
𝑓 ( ) = 2 ( ) 𝑒 −3∗3 = 𝑒 −2 = 2
3
3
3
3𝑒
2
4
27j) answer inflection point (3 , 3𝑒 2 )
27k) Sketch a graph
29) f(x) = ln(x-3)
29a) Find the domain
x–3>0
x>3
29a) answer (3, ∞)
29b) Find the x-intercept(s), if any
Loge(x-3) = 0
e0 = x – 3
1=x–3
4=x
29b) answer (4,0)
29c) Find the y-intercept, in there is one
f(0) = ln(0-3) = undefined
29c) answer none
29d) Find all vertical asymptotes and horizontal asymptotes
Ln graphs generally do not have horizontal asymptotes. I can see that it does not have a horizontal
asymptote once I sketch a graph on my calculator.
We find the vertical asymptote of the graph of a ln function by setting the argument equal to 0
Vertical asymptote computation
x–3=0
x=3
29d) answer vertical asymptote x = 3
no horizontal asymptote
29e) Find the interval(s) where the graph of the function is increasing
f’(x) =
1
𝑥−3
1=0
x–3 =0
No solution
x=3
(−∞, 3)
(3, ∞)
Ignore interval, not part of domain
test x = 4
f’(4) = 1
increasing
29e) answer increasing (3, ∞)
29f) Find the interval(s) where the graph of the function is decreasing
29f) answer never
29g) Find all relative maxima and relative minima
The graph never changes direction, there is no max or min
29g) answer none
29h) Find the interval(s) where the graph of the function is concave up (if any)
f’’(x) =
(𝑥−3)(0)−1(1)
(𝑥−3)2
−1
f’’(x) =(𝑥−3)2
-1 = 0
(x-3)2 = 0
No sol
x=3
(−∞, 3)
(3, ∞)
Ignore interval, not part of domain
test x = 4
f’’(4) = -1
concave down
29h) answer never concave up
28)i) Find the interval(s) where the graph of the function is concave down (if any)
29i) answer concave down (3, ∞)
29j) Find all inflection points (if any)
The graph doesn’t change concavity so there are no inflection points
29j) answer none
29k) Sketch a graph
31) f(x) = ln(2x+8)
31a) Find the domain
2x + 8 > 0
2x > -8
x > -4
31a) answer domain (−4, ∞)
31b) Find the x-intercept(s), if any
Loge(2x+8) = 0
e0 = 2x + 8
1 = 2x + 8
-7 = 2x
-7/2 = x
31b) answer x-intercept (-7/2, 0)
31c) Find the y-intercept,
f(0) = ln(2*0+8) = ln(8)
31c) answer y-intercept (0, ln(8))
31d) Find all vertical asymptotes and horizontal asymptotes
31d) answer
Ln graphs generally do not have horizontal asymptotes. I can see that it does not have a horizontal
asymptote once I sketch a graph on my calculator.
We find the vertical asymptote of the graph of a ln function by setting the argument equal to 0
Vertical asymptote computation
2x + 8 = 0
2x = -8
x = -4
29d) answer vertical asymptote x = -4
no horizontal asymptote
31e) Find the interval(s) where the graph of the function is increasing
f’(x) =
2
2𝑥+8
=
1
𝑥+4
1=0
x+4=0
No solution
x = -4
(−∞, −4)
(−4, ∞)
Ignore interval, not part of domain
test x = -3
f’(-3) = 1
increasing
31e) answer increasing (−4, ∞)
31f) Find the interval(s) where the graph of the function is decreasing
31f) answer never decreasing
31g) Find all relative maxima and relative minima
31g) answer none (as it never changes directions)
31h) Find the interval(s) where the graph of the function is concave up (if any)
f’’(x) =
(𝑥+4)(0)−1(1)
(𝑥+4)2
−1
f’’(x) = (𝑥+4)2
-1 = 0
(x+4)2 = 0
No solution
x+4=0
x = -4
(−∞, −4)
(−4, ∞)
Ignore interval, not part of domain
test x = -3
f’’(-3) = -1
concave down
31h) answer never concave up
31)i) Find the interval(s) where the graph of the function is concave down (if any)
31i) answer concave down (−4, ∞)
31j) Find all inflection points (if any)
The graph doesn’t change concavity so there are no inflection points
31j) answer none
31k) Sketch a graph
33) has been deleted