Hey guys,
Could i please have some clarification with the following question(s), working out by hand would be much appreciated, im pretty sure i have the right answers just want to double check 
Q1: Consider the function defined by f(x) = 2(x-3)^2.
a) Find the rule for the inverse.
Strictly speaking this question is ill-posed, because f(x) on its maximal domain does not have an inverse function, and no domain is initially given for which f(x) does have an inverse function.
 \Leftrightarrow x=f(y) \\<br />\begin{aligned}<br />x&=2(y-3)^2 \\<br />\frac{x}{2}&=(y-3)^2 \\<br />\sqrt{\frac{x}{2}}+3&=y\text{ If the domain of f is } [3,\infty) \\<br />-\sqrt{\frac{x}{2}}+3&=y\text{ If the domain of f is } (-\infty,3]<br />\end{aligned}<br />)
b) Restrict the domain of f to the form of [a, infinity] so that the inverse is also a function.
The graph of y=f(x) is a parabola with its turning point at x=3. Hence, any value of a ≥ 3 would make f a one-to-one function (and thus have an inverse). Choosing a = 3 is the largest domain that f can have such that it is one-to-one.
c) State the rules for the restricted f and f^-1 using function notation.
See answer to part (a)
d) Show that f(f^-1(x)) = x
^2 &= 2\left(\sqrt{\frac{x}{2}}\right)^2 \\<br />&= 2\frac{x}{2} \\<br />&= x<br />\end{aligned}<br />)
Q2: For: mx +2y = n and 3x + 6y = -1, find m and n for which the equations have:
a) a unique solution
b) an infinite number of solutions
b) no solution
Divide through the second equation by 3 to get a coefficient of 2 on y. This gives x + 2y = -1/3. Now, by inspection, the equations define the same line if m = 1 and n = -1/3, so this gives infinitely many solutions. The equations give parallel lines if m = 1 and n ≠ -1/3, so this gives no solution. The only other possibility is that m ≠ 1, in which case the lines are not parallel, so there is a unique solution.
For more challenging varieties of this kind of question (when there are unknown coefficients in BOTH equations), then you'll need to either (a) write both equations in the form ax + by = c, then use the determinant to find when there is no unique solution (ie. solve ad – bc = 0); OR (b) write both equations in the form y = mx + c, then equate the relevant parts and solve for when the lines coincide, are parallel, or intersect exactly once.
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