# The function is differentible in x=2. What is the function ? 2 eqns. for f f=x^2-1, x is < or equal to 2 f=mx+n if x>2

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Since the function is differentiable at `x = 2` , hence, the function needs to be continuous.

You should remember that a function is continuous if `lim_(x->2) f(2)` is finite such that:

`lim_(x->2,x=<2) (x^2-1) = lim_(x->2,x>2) (mx+n)`

Evaluating the left side limit yields:

`lim_(x->2,x=<2) (x^2-1) = 2^2 - 1 = 4 - 1 = 3`

`lim_(x->2,x>2) (mx+n) = 2m+ n`

Equating the result yields:

`2m + n = 3`

Notice that this equation is not helpful alone since it contains two unknowns, m and n, hence,you need one more equation.

Since the fuction is differentiable at `x=2` , then you should write following relations such that:

`{(f'(x) = 2x, x < 2),(f(x) = m, x>2):} => 2x = m`

Hence, substituting 2 for x, yields:

`4 = m`

You need to substitute 4 for m in the first equation such that:

`8+ n = 3 => n = 3 - 8 => n = -5`

**Hence, evaluating m and n, under the given conditions, yields `m = 4` and `n = -5.` **

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