Question

Suppose that the functions s and t are defined for all real numbers x as follows.s(x) = x-1t(x) = 3x - 5write the expressions for (s+t) (x) and (s, t)(x) and evaluate (s-t) (2).(s + 1)(x) = 1(5-1)(x) = 1(5-1) (2) = 0?

Answer 1

we are given the functions

`\begin{gathered} s(x)\text{ = x - 1} \\ t(x)\text{ = 3x - 5} \end{gathered}`

1. we are to find (s + t)(x)

given s(x) = x -1 and t(x) = 3x - 5 we have

`\begin{gathered} (s+t)(x)\text{ = x -1 + 3x - 5} \\ (s+t)(x)=4x\text{ - 6} \end{gathered}`

Therefore, (s + t)(x) = 4x - 6

2. we are to find (s - t)(x)

using the functions s(x) and t(x), this will give

`\begin{gathered} (s-t)(x)\text{ =s(x) - t(x)} \\ (s-t)(x)=x-1-(3x-5) \\ (s-t)(x)=x-1-3x+5 \\ (s-t)(x)=-2x+4 \end{gathered}`

Therefore, (s - t)(x) = -2x + 4

3. we are to find (s.t)(2)

first we need to find

(s.t)(x)

using the functions s(x) and t(x) we have

`\begin{gathered} (s\mathrm{}t)(x)=s(x).t(x) \\ (s\mathrm{}t)(x)=(x-1)(3x-5) \end{gathered}`

Therefore, (s.t)(x) = (x - 1)(3x - 5)

Hence

`\begin{gathered} (s\mathrm{}t)(2)\text{ }\Rightarrow\text{ x = 2} \\ \text{therefore} \\ (s\mathrm{}t)(2)=(2-1)(3(2)-5) \\ (s\mathrm{}t)(2)=(1)(6-5) \\ (s\mathrm{}t)(2)=(1)(1) \\ (s\mathrm{}t)(2)=1 \end{gathered}`

Therefore, (s.t)(2) = 1