Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0
Answer:
A = 9
B = 2
C = 5
D = 1
E = 8
F = 7
Step-by-step explanation:
A 9 B 2 C 5 = 16
D 1 E 8 F 7 = 16
10 10 12
BxC (2x5) = 10
A+D (9+1) = 10
Well to answer question 43, you would have to multiply the radius but 2 so 8*2= 16 and 16* 3.14= 50.24
ANSWER: 50.24
Step-by-step explanation:
zidufhehkcsyjodbjd r
Answer:
y=2/3x
Step-by-step explanation:
2/3 is greater than 1/2. It doesn't matter if one of the slopes are negative.