1a) Find the set of values of x for which 0 < ( 2 - x - x^2 ) / x^2 < 2
1b) Find the cordinates of the turning point on the curve y = ( 2 - x - x^2 ) / x2
Mr F says: Solve 
. You can get dy/dx using either the quotient rule or seeing that 
.
2) sketch y = root of (x-2)
Mr F says: Upper half of sideways parabola. Vertex at (2, 0).
3) Find x such that:
a) x^4 - 5x^2 6 > 0
Mr F says: Do you mean x^4 - 5x^2 + 6 > 0 or x^4 - 5x^2 - 6 > 0? Either way, I suggest drawing a graph and looking for the values of x for which the graph is above the x-axis.
b) | 3 - 2x < or equal to | x + 4 |
Mr F says: You might start by drawing graphs of y = |3 - 2x| = |2x - 3| and y = |x + 4|. Then find the x-coordinates of the intersection points ....
4a) show y = x - 6 is an asymptotoe of the curve y = (x-1)(x-3)/(x+2)
Mr F says:
using polynomial long division .....
b) Sketch the curve showing cleanly the other asumptote, the points of intersection of the curve with the co-ordinate axes.
Mr F says: Vertical asymptote x = -2. You should know how to find x- and y-intercepts (especially y-intercept!!).
PLEASE PLEASE PLEASE HELP
LinkBack URL
About LinkBacks
Originally Posted by ziad123
1a) Find the set of values of x for which 0 < ( 2 - x - x^2 ) / x^2 < 2
