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
The rays shown in the pictur: Ray VX, YZ , XY, ZV..... Plus more!
Answer:
4x+4
Step-by-step explanation:
Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identities
secx =
, cosecx = 
cotx =
, tanx = 
Consider the left side
secA cosecA - cotA
=
×
- 
=
- 
= 
=
( cancel sinA on numerator/ denominator )
= 
= tanA = right side ⇒ proven
1. 8b+40
2. -36n+16
3. -6y-3
4. 4k+32
5. 3n^2 (squared)