Question

Please help struggling in math & this is for practice

Answer 1

The given functions are

`\begin{gathered} f(t)=(11)/(t-2) \\ g(t)=(22t)/(t^2-4) \end{gathered}`

The first part consists in adding the functions f(t) + g(t)

`f(t)+g(t)=(11)/(t-2)+(22t)/(t^2-4)=(11(t^2-4)+22t(t-2))/((t-2)(t^2-4))`

Let's factor the square difference

`(11(t+2)(t-2)+22t(t-2))/((t-2)(t^2-4))=(11(t+2)+22t)/(t^2-4)`

Now, we simplify

`(11t+22+22t)/(t^2-4)=(33t+22)/(t^2-4)`

Hence, the addition of the given functions is

`f(t)+g(t)=(33t+22)/(t^2-4)`

On other hand, let's solve the difference

`\begin{gathered} f(t)-g(t)=(11)/(t-2)-(22t)/(t^2-4)=(11(t^2-4)-22t(t-2))/((t-2)(t^2-4)) \\ f(t)-g(t)=(11(t+2)(t-2)-22t(t-2))/((t-2)(t^2-4)) \\ f(t)-g(t)=(11(t+2)-22t)/(t^2-4)=(11t+2-22t)/(t^2-4) \end{gathered}`

Hence, the difference is

`f(t)-g(t)=(-11t+2)/(t^2-4)`