Question

Determine the value of the constant a if the function `f(x)` defined below is continuous at `x=2`. `f(x)={(ax^2+7x; x≤2),(3x^2+3a; x>2):}` ?

Answer 1

Please see below.

Explanation:

`f` is continuous at `a` if and only if `f(a) = lim_(xrarra)f(x)`.

In order for `f` to be continuous at `2`, we need `lim_(xrarr2)f(x)` to exist and to be equal to `f(2)`.

This function changes rules at `2`, so to make sure that the limit at `2` exists, we need the left and right limits at `2` to exist and to be equal.

So, find `lim_(xrarr2^-)f(x)` and find `lim_(xrarr2^+)f(x)`

Set those equal to each other and solve for `a`.

Check the `f(2)` gives the same value as the limit.

You should get `a = -2`

Answer 2

`a = -2`

Explanation:

If the piecewise function is continuous at `x =2`, then the pieces must be the same value at `x =2`, and we can, therefore, set them equal to each other:

`a(2^2)+ 7(2) = 3(2^2)+ 3a`

`4a+ 14 = 12+ 3a`

`a = -2`

`f(x)={(-2x^2+7x; x≤2),(3x^2-6; x>2):}`

Here is a graph of `f(x)`:

Please observe that the graph is continuous.

Answer 3

`-2`.

Explanation:

For the given fun. `f` to be continuous at `x=2`, we must have,

`lim_(x to 2-)f(x)=f(2)=lim_(x to 2+)f(x)............(square)`.

Now, as `x to 2-, x lt 2. :. f(x)=ax^2+7x`.

`:. lim_(x to 2-)f(x)=lim_(x to 2-)ax^2+7x=a*2^2+7*2`.

`:. lim_(x to 2-)f(x)=4a+14......................(square1)`.

Similarly, `lim_(x to 2+)3x^2+3a=3*2^2+3*a`.

`:. lim_(x to2+)f(x)=3a+12......................(square2)`.

Then, from `(square), (square1) and (square2)`, we have,

`4a+14=3a+12," giving, "a=-2`.