# Functions

• Aug 7th 2010, 04:02 AM
rcs
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
• Aug 7th 2010, 04:22 AM
sa-ri-ga-ma
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)]
• Aug 7th 2010, 06:01 AM
eumyang
Quote:

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?

Quote:

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.
• Aug 7th 2010, 06:44 AM
Soroban
Hello, rcs!

Quote:

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

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

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

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

• Aug 7th 2010, 07:19 AM
Soroban
Hello again, rcs!

Another approach to the first one . . .

Quote:

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

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

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

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

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

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

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

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

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 $x\!:$

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

• Aug 8th 2010, 05:50 PM
rcs
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
• Aug 8th 2010, 05:59 PM
eumyang