Question

2. The functionsf and g are defined as below. f(x) = 3x+2: XER g(x)= 6 2x +3 Find the value of x for which f(g(x)) = 3 Sketch in a single diagram, the graphs of f(x) and f(x). Express each of f(x) and g(x), and solve the equation f¹(x) = g(x)

Answer 1

$$f(x)=3x+2, x\in \R$$

$$g(x)=\dfrac{6}{2x+3}, x\in \R, x\not=-1.5$$

(i)

$$f(g(x)) =3(\dfrac{6}{2x+3})+2$$

Given

$$f(g(x)) =3$$

$$3(\dfrac{6}{2x+3})+2=3$$

$$\dfrac{18}{2x+3}=1$$

$$2x+3=18$$

$$2x=15$$

$$x=7.5$$

(ii)

$$f(x)=3x+2$$

Replace $f(x)$ with $y$

$$y=3x+2$$

Interchange $x$ and $y$

$$x=3y+2$$

Solve for $y$

$$y=\dfrac{1}{3}x-\dfrac{2}{3}$$

Replace $y$ with $f^{-1}(x)$

$$f^{-1}(x)=\dfrac{1}{3}x-\dfrac{2}{3}$$

The graph of a function and its inverse are symmetric with respect to the line $y=x.$

(iii)

$$g(x)=\dfrac{6}{2x+3}, x\in \R, x\not=-1.5$$

Replace $g(x)$ with $y$

$$y=\dfrac{6}{2x+3}$$

Interchange $x$ and $y$

$$x=\dfrac{6}{2y+3}$$

Solve for $y$

$$2y+3=\dfrac{6}{x}$$

$$2y=\dfrac{6}{x}-3$$

$$y=\dfrac{3}{x}-1.5$$

Replace $y$ with $g^{-1}(x)$

$$g^{-1}(x)=\dfrac{3}{x}-1.5, x\not=0$$

Given

$$f^{-1}(x)=g^{-1}(x)$$

$$\dfrac{1}{3}x-\dfrac{2}{3}=\dfrac{3}{x}-1.5$$

$$x^2-2x=9-4.5x$$

$$x^2+2.5x-9=0$$

$$D=(2.5)^2-4(1)(-9)=42.25$$

$$x=\dfrac{-2.5\pm\sqrt{42.25}}{2(1)}=-1.25\pm3.25$$

$$x_1=-1.25-3.25=-4.5, x_2=-1.25+3.25=2$$

$$x\in\{-4.5, 2\}$$