Question

g(x) = ax + b / x, x≠k

given g(2)-3 & g(-2)=5

a) value of k

b) the values of a and b

c) the value of h if g-1(h)=2 (is this inverse)?

Answer 1

Solution.

$$g(x)=\frac{ax+b}{x}, x\neq k$$

a) such as domain g(x) is equal $(-\infty,0)\bigcup (0,\infty),$ so $k=0.$

b) such as g(2)=-3, we will have $\frac{2a+b}{2}=-3.$ From here $2a+b=-6.$

such as g(-2)=5, we will have $\frac{-2a+b}{-2}=5.$ From here $-2a+b=-10.$

Solve system

$-2a+b=-10,\newline 2a+b=-6.$

Solve this system by adding:$-2a+2a+b+b=-10-6,\newline 2b=-16,\newline b=-16:2,\newline b=-8.$

Substitute the value of b in the first equation of the system, then

$-2a-8=-10,\newline -2a=-2,\newline a=1.$

So, $a=1, b=-8.$

c) $g(x)=\frac{x-8}{x}.$

Find $g^{-1}(x).$ To do this, express the variable x through the variable g:

$gx=x-8,\newline gx-x=-8,\newline (g-1)x=-8,\newline x=\frac{-8}{g-1}=\frac{8}{1-g}.$

And swap the variables g and x. We will have $g^{-1}(x)=\frac{8}{1-x}.$

Such as $g^{-1}(h)=2,$ so

$\frac{8}{1-h}=2.$

From here $1-h=4,$ and $h=-3.$