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 $\displaystyle {\color{red}\frac{dy}{dx} = 0}$. You can get dy/dx using either the quotient rule or seeing that $\displaystyle {\color{red}y = \frac{2}{x^2} - \frac{x}{x^2} - \frac{x^2}{x^2} = 2 x^{-2} - x^{-1} - 1}$.
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:
$\displaystyle {\color{red} y = \frac{x^2 - 4x + 3}{x+2} = x - 6 + \frac{15}{x+2}}$ 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
Originally Posted by ziad123
1a) Find the set of values of x for which 0 < ( 2 - x - x^2 ) / x^2 < 2
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