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:
9
Step-by-step explanation:
20 x 5 = 100
45/5 = 9
9 to 20 = 11
45 to 100 = 55
<u>55</u> = 5<--- Linear
11
(since the total answer is linear, the answer is right)
Answer:
No, it is not prime.
Step-by-step explanation:
67^2 = 4489
A number cannot be prime if it is divisible by a number besides 1 and itself. 4489 is divisible by 67; no perfect square can be prime.
Answer:
2x*x
2x*-5
1*x
1*-5
2x²-10x+x-10 add numbers that have similar variables
-2x²-10x+x-10
=[-2x²-9x-10]