1. Functions

1. if f(g(1/x)) = (x+2)x and f(x) and f(x)= 1/(3-x), find g(x).

2. Find the equation of the line through the origin and bisecting the angle between the axes in the first and third quadrants.

3. Find a so that the system below has an infinite number of solutions
ax + (2/3)y = 1/2
2x + (3/2)y = 9/8

2. f(g(1/x)) = 1/[3 - g(1/x)] = (x+2)x

3 - g(1/x) = 1/[(x+2)x]

g(1/x) = 3 - 1/[(2x+2)x]

g(x) = 3 - 1/[(2/x +2)(1/x)]

3. Originally Posted by rcs
2. Find the equation of the line through the origin and bisecting the angle between the axes in the first and third quadrants.
Wouldn't this just be y = x? Or am I not reading this right?

3. Find a so that the system below has an infinite number of solutions
ax + (2/3)y = 1/2
2x + (3/2)y = 9/8
For this system to have an infinite number of solutions one equation has to be a "multiple" of another. Notice that (2/3)(9/4) = 3/2 and (1/2)(9/4) = 9/8. So a * (9/4) = 2. Solve for a.

4. Hello, rcs!

$\text{3. Find }a\text{ so that this system has an infinite number of solutions:}$

. . . . $\begin{array}{cccc}ax + \frac{2}{3}y &=& \frac{1}{2} & [1] \\ \\[-3mm]
2x + \frac{3}{2}y &=& \frac{9}{8} & [2] \end{array}$

$\begin{array}{ccccccc}
\text{Multiply [2] by -}\frac{4}{9}: & \text{-}\frac{8}{9}x - \frac{2}{3}y &=& \text{-}\frac{1}{2} \\ \\[-4mm]
\text{Add [1]:} & ax + \frac{2}{3}y &=& \frac{1}{2} \end{array}$

And we have: . $ax - \frac{8}{9}x \:=\:0 \quad\Rightarrow\quad \left(a-\frac{8}{9}\right)x \;=\;0$

Evidently, the answer is: . $x \,=\,0$

But if $(a-\frac{8}{9})$ equals 0, $x$ can be any value.

Therefore: . $a - \frac{8}{9} \:=\:0 \quad\Rightarrow\quad \boxed{a \:=\:\tfrac{8}{9}}$

5. Hello again, rcs!

Another approach to the first one . . .

$\text{1. if }f\!\left(g(\frac{1}{x})\right) \:=\: x(x+2)\:\text{ and }\:f(x)\:=\: \dfrac{1}{3-x}$

. . , $\text{find }\:g(x).$

$f\left(g(\frac{1}{x})\right) \;=\;\dfrac{1}{3-g(\frac{1}{x})} \;=\;x(x+2)$

. . . . . . $\bigg[3-g(\frac{1}{x})\bigg]\bigg[x(x+2)\bigg] \;=\;1$

. . . $3x(x+2) - x(x+2)\,g(\frac{1}{x}) \;=\;1$

. . . . . . . . . . . $x(x+2)\,g(\frac{1}{x}) \;=\;3x(x+2)-1$

. . . . . . . . . . . $x(x+2)\,g(\frac{1}{x}) \;=\;3x^2 + 6x + 1$

. . . . . . . . . . . . . . . . . $g(\frac{1}{x}) \;=\;\dfrac{3x^2+6x+1}{x^2+2x}$

Divide numerator and denominator by $x^2\!:$

. . $g(\frac{1}{x}) \;=\;\dfrac{3 + \frac{6}{x} - \frac{1}{x^2}}{1 + \frac{2}{x}} \;=\;\dfrac{3 + 6\left(\frac{1}{x}) - (\frac{1}{x})^2}{1 + 2(\frac{1}{x})}$

Replace $\frac{1}{x}$ with $x\!:$

. . $g(x) \;=\;\dfrac{3 + 6x + x^2}{1 + 2x}$

6. i thank you so much for your help... you are all my angels in my field... thank u guys. i hope somebody could help me out for my # 2 problems... thank you mathematicians . but please let me know how to type the mathematical operators/symbols in this page.. let me know how... please.

rcs

8. thank you eumyang ... i gues y = x ryt. sorry im a little engrossed with my assignment i cud not think wisely ... even simple problems i could think they are already so hard sorry