Question

suppose that the functions s and t are defined for all real numbers x as follows.

s(x)=2x + 5
t(x)=x - 6

write the expressions for ((s + t)(x)) and ((scdot t)(x)) and evaluate ((s - t)(1)).

((s + t)(x)=3x - 1)
((scdot t)(x)=2x^{2}-30)
((s - t)(1)=0

Explanation:

Step1: Find \((s + t)(x)\)

Add the two functions: \((s + t)(x)=s(x)+t(x)=(2x + 5)+(x - 6)=2x+x+5 - 6=3x - 1\)

Step2: Find \((s\cdot t)(x)\)

Multiply the two functions: \((s\cdot t)(x)=s(x)\cdot t(x)=(2x + 5)(x - 6)=2x\cdot x-2x\cdot6+5\cdot x - 5\cdot6=2x^{2}-12x + 5x-30=2x^{2}-7x - 30\)

Step3: Find \((s - t)(1)\)

First, find \((s - t)(x)=s(x)-t(x)=(2x + 5)-(x - 6)=2x + 5-x + 6=x + 11\). Then substitute \(x = 1\) into \((s - t)(x)\), so \((s - t)(1)=1+11 = 12\)

Answer:

\((s + t)(x)=3x - 1\)
\((s\cdot t)(x)=2x^{2}-7x - 30\)
\((s - t)(1)=12\)