Question

Find the values of m and b that make f continuous everywhere: m = ? b = ?

`f(x) = {((7+6x-x^2)/(x+1)",", x < -1),(mx+b",", x ∈ [-1,4]),(2*2^(4-x)+16",", x>4):}`

Have never seen a piecewise function quite like this before and not sure how to even graph the mx+b line in all honesty.

Answer 1

`{(m=2),(b=10):}`

Explanation:

We can see that each individual function will be continuous on their domains.

To ensure that the function is continuous, we have to find the values of `m` and `b` that make the values of the functions equal at `x=-1` and `x=4`, where the piecewise switches from one function to another.

First looking at `x=-1`, we have to make `(7+6x-x^2)/(x+1)` and `mx+b` equal at `x=-1`. If they have the same value, then the function is continuous there.

At `x=-1`, we see that `(7+6x-x^2)/(x+1)` is undefined, but we can find the limit at `-1` by factoring the numerator:

`lim_(xrarr-1)(7+6x-x^2)/(x+1)=lim_(xrarr-1)((7-x)(1+x))/(x+1)=lim_(xrarr-1)(7-x)=8`

So, we need `mx+b` to equal `8` at `x=-1`. That is, `-m+b=8`.

We can't solve explicitly for `m` or `b` yet, so we also have to find a relation ensuring continuity at `x=4`, where the function switches from `mx+b` to `2*2^(4-x)+16`.

At `x=4`, we see that `mx+b` is equal to `4m+b` and the other function is `2*2^0+16=18`. Then, `4m+b=18`.

So, we have the two relations of `m` and `b`:

`{(-m+b=8),(4m+b=18):}`

Solving this gives:

`{(m=2),(b=10):}`