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
Step-by-step explanation:
2(3x-5)=5x-3
6x -10 =5x -3
x=7
Answer:
i believe it would be 6 1/2 simplified or 6 2/4 unsimplified
Step-by-step explanation:
Answer:
5 1/6
Step-by-step explanation:
Rewriting our equation with parts separated
=8+2/4−3−1/3
Solving the whole number parts
8−3=5
Solving the fraction parts
2/4−1/3=?
Find the LCD of 2/4 and 1/3 and rewrite to solve with the equivalent fractions.
LCD = 12
6/12−4/12=2/12
Reducing the fraction part, 2/12,
2/12=1/6
Combining the whole and fraction parts
5+1/6=5 1/6