Question

For which value(s) of k,is the function f,defined as below,continuous at x=2?f(x)={3-kx,1≤x<2 and x²/4-3, x≥2.Further,at which other points in [1,infinity[ is f continuous,and why?

Answer 1

`k=5/2`

Explanation:

Let:

`f(x) = {(3-kx " for " 1 <= x < 2),(x^2/4-3 " for " x >=2):}`

Note that in the interval `x in (1,2)` we have:

`f(x) = 3-kx`

and in the interval `x in (2,+oo)`

`f(x) = x^2/4-3`

so in the interior of each interval `f(x)` equals a polynomial function and is therefore continuous.

Evaluate now the left and right limit of `f(x)` for `x->2`:

`lim_(x->2^-) f(x) = lim_(x->2^-) 3-kx = 3-2k`

`lim_(x->2^+) f(x) = lim_(x->2^+) x^2/4-3 = -2`

In order for `f(x)` to be continuous for `x=2`, in that point the function must have a finite limit and this limit must be equal to `f(2)`.

For the limit to exist, the left and right limits must coincide, so:

`3-2k = -2`

so:

`k=5/2`

and in such case:

`lim_(x->2) f(x) = -2 = f(2)`.