Rewrite the given equation as
.
Since
, you can conclude that the imimum value of y will be gained for the minimum value of
. The minimum value of
is 0 for x=2.
So, y(2)=0+4=4.
Answer: minimum value of y is 4, when x=2.
Answer:
Direct & Inverse Proportion (H) - Version 2 January 2016 . A collection of 9-1 Maths GCSE Sample and Specimen questions from AQA, . 1. 2. Write an expression for y in terms of x. [4]. 2. A pebble is thrown vertically upwards. . (b) Find the initial speed of the pebble if the maximum height reached is 16 m. . T is given by.
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
