The height at which the bird flies (y) is related to the distance from its nest by y = 3x^3 + 2x^2 - 10x + 5.
The graph of the relation is:
The distance of the bird from its nest is a positive number (as this is the magnitude of a vector). From the graph it is evident that there is no limit to how large y can be for positive values of x.
To determine the lowest value of y, calculus can be used to determine y' and y''. Solve y' = 0 for x
y' = 9x^2 + 4x - 10
y' = 0
=> 9x^2 + 4x - 10 = 0
The positive root of the equation is `(sqrt 94 - 2)/9` . At this point the value of y is negative and equal to -0.2128.
But the height of a bird (using logic) cannot be negative, it cannot fly below the surface.
In case the height can take on negative values, the minimum height of the bird is -0.2128
The height y, refers to height of the bird from the tree (hopefully the tree-top). Therefore, it can take on negative values (of course in the range of the actual height of the tree). The bird can definitely fly at a level, a bit lower than the tree-top.
The minimum height of the bird (with respect to the tree-top) is thus -0.2128 m.
With help of calculus we can define minimum and maximum local max or local min . Some time global max /min. Here global max /min does not exist . Local min /max exist.But it is physical problem so non negative restriction will effect the results.It is simple to observe from graph and then logical conclusion. Let first draw the graph
when y=0 ,possible distance of the bird is x=1 , x=.61 and x=-2.28
distance can not be negative so neglect the value of x=-2.28
Thus when x=1 or .61 ,y=0
Thus we conclude that minimum height of the bird is zero.