Question

Show that f(x)=2a+3b is continous, where a and b are constants?

Answer 1

See explanation below

Explanation:

A function is called continuous in `x_0` if

`f(x_0)` exists and is finite

`lim_(xtox_0)f(x)` exists and finally

`lim_(xtox_0)f(x)=f(x_0)`

In our case `f(x)=2a+3b` with `a` and `b` constants. So the function doesn't depend of x and for this reason is constant for every value of x, and his value is `f(x)=2a+3b` whatever be a and b.

For this reason

`f(x_0)=2a+3b` exists for every `x_0` and is finite
`lim_(xtox_0)f(x)=2a+3b`. This proof that `f(x)` is continous in all `x in RR`