Accedmychevron_right11thchevron_rightmathchevron_rightFunctions And Graphschevron_rightExercise 2.1

Solution

Given f(x)=3x5f(x)=3x-5.

(i) Domain and range

A linear function is defined for all real xx and produces all real values.

Df=R,Rf=R\boxed{D_f=\mathbb{R},\quad R_f=\mathbb{R}}

(ii) One-to-one

Assume f(x1)=f(x2)f(x_1)=f(x_2):

3x15=3x253x1=3x2x1=x23x_1-5=3x_2-5\Rightarrow 3x_1=3x_2\Rightarrow x_1=x_2

So ff is one-to-one.

f is one-to-one\boxed{f\ \text{is one-to-one}}

(iii) Onto (co-domain R\mathbb{R})

Let yRy\in\mathbb{R}. Solve y=3x5y=3x-5:

y+5=3xx=y+53y+5=3x\Rightarrow x=\frac{y+5}{3}

Since y+53R\frac{y+5}{3}\in\mathbb{R}, every real yy has a preimage, so ff is onto.

f is onto R\boxed{f\ \text{is onto }\mathbb{R}}