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.

1 Answer
Apr 16, 2017

Answer:

#{(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):}#