Find the concavity changes of f(x).
Give the x-coordinates.
So, the answer to the multiple choice question is -1 and 5 since those points are where f''(x) changes sign.
Note: You can tell that you're supposed to find the concavity changes of f(x) and not f''(x) because from the graph it's obvious that the concavity of f"(x) doesn't change.
- Given, x > 0, y > 0.
- So, the values of x and y are less than 0, that means, they are negative integers.
- If we divide a Cartesian plane with the x-axis and y-axis, we get four quadrants.
- 1st Quadrant : (+,+)
- 2nd Quadrant : (-,+)
- 3rd Quadrant : (-,-)
- 4th Quadrant : (+,-)
- Since the values of x and y are both negative, so (x, y) is located in the <em><u>3rd Quadrant</u></em>.
Hope you could get an idea from here.
Doubt clarification - use comment section
Answer:
4 hours
Step-by-step explanation:
72+12x=120 and when you isolate x you get x=4
Answer:
we get:
= 1/2 ʃ02π ʃ14 ( √u -1)du dθ
= ʃ02π 11/12 dθ
= 11/6 π
Step-by-step explanation:
Given:
Radius = 2 units
Plane 1 unit from center of sphere.
The volume of D as an triple integral in spherical, cylindrical and rectangular coordinates are:
Spherical:
ʃ02π ʃ0π/3 ʃsecΦ2 p2 sinΦ dp dΦ dθ
Cylindrical:
ʃ02π ʃ0√3 ʃ1√4-r2 r dz dr dθ
Rectangular:
ʃ-√3√3 ʃ-√3-x2√3-x2 ʃ1√4-x2-y2 1dz dy dx
Solving the integral by using cylindrical coordinates:
ʃ02π ʃ0√3 ʃ1√4-r2 r dz dr dθ = ʃ02π ʃ0√3 r ( √(4-r2) -1) dr dθ
put u = 4-r2, by substituting,
we get:
= 1/2 ʃ02π ʃ14 ( √u -1)du dθ
= ʃ02π 11/12 dθ
= 11/6 π
