1.Consider the function f(x)= (3/4)x4 - x3 - 3x2 +6x
A. Find the relative extrema for f(x); be sure to label each as a maximum or minimum. You do not need to find function values; just find the x-values.
B. Determine the interval(s) where f(x) is increasing (if any) and the interval(s) where f(x) is decreasing (if any).
C. Show why there are exactly two inflection points for this function. Note: You do NOT need to find the inflection points.
D. Show f(x) is concave up at x = −2, x = −1, and x = 2. And show f(x) is concave down at x = 0. Use this information and the information in parts A–C to sketch this function.
2. A machine starts dumping sand at the rate of 20 m3/min, forming a pile in the shape of a cone. The height of the pile is always twice the length of the base diameter. The volume formula for a right circular cone is V= 1/3pi*r2.*h
A. After 5 minutes, what is the height of the pile?
B. After 5 minutes, how fast is the height increasing?
C. After 5 minutes, how fast is the base radius increasing?
D. After 5 minutes, how fast is the area of the base increasing?
#1 Extrema occur where the derivative is zero
f(x)= (3/4)x^4 - x^3 - 3x^2 +6x
f'(x) = 3x^3 - 3x^2 - 6x + 6
You can factor that by grouping:
3x^2(x-1) - 6(x-1)
(3x^2-6)(x-1)
3(x^2-2)(x-1)
so the extrema occur at x = ±√2, 1
f(x) is increasing where f'(x) > 0
so just figure that out.
f"(x) = 9x^2 - 6x - 6 = 3(3x^2-2x-2)
Since the discriminant is positive, f" has two zeroes, so f has two inflection points
f is concave up where f" > 0
type in your function(s) at wolframalpha.com to help you see the graph(s)
#2
since h = 2r, V = π/3 r^2 h = 2π/3 r^3 or π/3 (h/2)^2 h = π/12 h^3
(A) at t=5, π/12 h^3 = 100
so, h = ∛(1200/π) m
(B) dV/dt = π/4 h^2 dh/dt
20 = π/4 ∛(1200/π)^2 dh/dt
so, dh/dt = 20 * 4/π * ∛(π/1200)^2 = 1/∛(180π) m/min
(C) since h = 2r, dh/dt = 2 dr/dt
(D) since the base area is A=πr^2,
dA/dt = 2πr dr/dt = 2π * 1/2 * ∛(1200/π) * 1/(2∛(180π)) = ∛(5π/6)
check my arithmetic. The calculus is correct, I'm sure.
Did you get f ' (x) = 3x^3 - 3x^2 - 6x + 6 for the derivative.
(if you didn't you have a very serious problem in Calculus)
you MUST know that for relative extrema , f ' (x) = 0 , so
3x^3 - 3x^2 - 6x + 6 = 0
x^3 - x^2 - 2x + 2 = 0
by grouping:
x^2(x - 1) - 2(x - 1) = 0
take over, you clearly will get 3 distinct values of x
b) You MUST also know that the original function is increasing when f ' (x) is positive
and decreasing when f ' (x) is negative
It will help to sketch the graph the derivative.
c) you MUST also know that at any point of inflection
f '' (x) = 0
so your f '' (x) = 9x^2 - 6x - 6
So set this equal to zero, and you have a quadratic in x
which will give you two distinct solutions
d) if the function's graph is concave up , then f '' (x) > 0
for concave down, f '' (x) < 0
sub in the given values and decide.
#2, show me what you have done so far. Did you make a sketch?
What are your answers?
Unfortunately, I am not quite sure how to solve these
1.
A) 3x^3 - 3x^2 - 6x + 6 = 0
x^3 - x^2 - 2x + 2 = 0
x^2(x - 1) - 2(x - 1) = 0
(x^2-2)(x-1) = 0
x^2-2 = 0
x^2 = 2
x = ±√2
x-1 = 0
x = 1
The extrema = ±√2 (1.4) and 1
B) f’(x) = 3x^3 - 3x^2 - 6x + 6
f’(1.2) = 3(1.2)^3 - 3(1.2)^2 - 6(1.2) + 6
f’(1.2) = 3(1.728) - 3(1.44) - 7.2 + 6
f’(1.2) = 5.184 - 4.32 - 1.2
f’(1.2) = -0.336
f’(x) = 3x^3 - 3x^2 - 6x + 6
f’(-1.2) = 3(-1.2)^3 - 3(-1.2)^2 - 6(-1.2) + 6
f’(-1.2) = -5.184 - 4.32 + 13.2
f’(-1.2) = 3.696
f’(x) = 3x^3 - 3x^2 - 6x + 6
f’(-1.5) = 3(-1.5)^3 - 3(-1.5)^2 - 6(-1.5) + 6
f’(-1.5) = -10.125 - 6.75 + 15
f’(-1.5) = -1.875
f’(x) = 3x^3 - 3x^2 - 6x + 6
f’(1.5) = 3(1.5)^3 - 3(1.5)^2 - 6(1.5) + 6
f’(1.5) = 10.125 - 6.75 - 3
f’(1.5) = 0.375
- |-√2| + |1| - |√2| +
The graph is increasing from -√2 to 1 and √2 to infinity. The graph in decreasing from -infinity to -√2 and 1 to √2
C) f '' (x) = 9x^2 - 6x - 6
(9x^2 - 6x - 6)/3 = 0/3
3x^2 - 2x - 2 = 0
x= (-b ± √(b^2 - 4ac))/2a
x= (-2 ± √(-2^2 - 4(3)(-2)))/2(3)
x= (-2 ± √(4 + 24))/9
x= (-2 ± √28)/9
x= (-2 - √28)/(9)
x= (-2 + √28)/9
D) Concave up: x = −2, x = −1, and x = 2. Concave down at x = 0
concave up: f '' (x) > 0
concave down: f '' (x) < 0
f '' (0) = 9x^2 - 6x - 6
f '' (0) = 9(0)^2 - 6(0) - 6
f '' (0) = -6
f '' (0) < 0 so it is concave down.
f '' (-1) = 9x^2 - 6x - 6
f '' (-1) = 9(-1)^2 - 6(-1) - 6
f '' (-1) = 9 + 6 - 6
f '' (-1) = 9
f '' (-1) > 0 so it is concave up.
f '' (2) = 9x^2 - 6x - 6
f '' (2) = 9(2)^2 - 6(2) - 6
f '' (2) = 36 -12 - 6
f '' (2) = 18
f '' (2) > 0 so it is concave up.
f '' (-2) = 9(-2)^2 - 6(-2) - 6
f '' (-2) = 36 + 12 - 6
f '' (-2) = 42
f '' (-2) > 0 so it is concave up.
2:
A) h = 2d or 4r
v = π/3 r^2 h
v = π/3 (h/4)^2 h
v = π/48 h^3
After 5 minutes, 100m^3 of sand would have been dumped. Therefore:
π/48 h^3 = 100
So 100 = π/48 h^3
48 (100) = (π/48 h^3) 48
4800 = π*h^3
4800/π = h^3
∛(4800/π) = h
approximately 11.51m
B) dv/dt = (?)
C) since h = 4r, dh/dt = 4 dr/dt (?)
D) base area: A=πr^2,
dA/dt = 2πr dr/dt
2π * (∛(4800/π))/4 * dr/dt
2π ∛(4800/4π) * dr/dt
(I’m a bit stuck on how to continue solving this ;;;)
Graphs:
cdn.discordapp.com/attachments/328266521849888769/523338050890366976/unknown.png
cdn.discordapp.com/attachments/328266521849888769/523311358066950167/unknown.png
if v = π/48 h^3
then dv/dt = π/16 h^2 dh/dt
since dv/dt = 20, plug that in and solve for dh/dt
then, recall that since h=2r, dh/dt = 2 dr/dt