We're asked to determine the values of a and b
that will make the function continuous
for all real numbers.
We have a piecewise function with three pieces.
If we take a look at the graphs
of these two piecewise functions,
notice on the left,
we have a continuous piecewise function.
We can sketch the function without lifting up our pencil,
meaning the graph has no holes or breaks.
However on the right,
we have a piecewise function with discontinuity,
because there are breaks or holes in the graph.
Notice how there's a break at x equals negative two,
as well as x equals positive two.
So by analyzing these two graphs,
notice how for a piecewise function to be continuous,
the function values or y values must be the same,
where one piece stops and the next piece starts.
For example on the left,
this occurs at x equals negative two,
as well as x equals positive two.
Even though one piece may end with an open point,
the other piece starts with a closed point,
making the graph continuous.
So going back to our example,
this indicates that 2 times e to the power of bx plus 3a
must equal ax plus b at x equals -1,
and ax plus b must equal b times natural log of the quantity
x plus 1, plus 2 at x equals 0,
in order for the piecewise function to be continuous.
So let's go ahead and set this up
using the first and second function rule.
2 times e to the power of bx plus 3a
must equal ax plus b
at x equals -1
and ax plus b
must equal b natural log of the quantity
x plus 1, plus 2
at x equals 0.
If these two conditions are met,
the piecewise function will be continuous.
So let's begin with this first condition
and substitute -1 for x.
This gives us 2 times e to the power of b times -1
plus 3a equals a times -1 plus b.
And now let's simplify.
We have 2, e to the power of -b
plus 3a equals -a plus b.
Let's combine the a terms
by adding a to both sides of the equation.
Simplifying, we have 2 times e to the power of -b
plus 4a equals b.
Notice this equation contains two unknowns,
so we'll have to stop here
and hopefully get some more information
from the second equation.
So going to the second condition,
we will substitute 0 for x,
which gives us a times 0 plus b
equals b times natural log of the quantity
0 plus 1, plus 2.
Simplifying, of course, a times 0 is 0,
giving us b equals.
On the right, notice we have natural log one,
well natural log one is equal to zero.
For quick review, remember natural log is log base e.
The reason natural log one is zero
is because e to the power of zero is equal to one.
So if natural log one is zero,
we have b times zero, which is zero,
and we're left with b equals 2.
Well now that we know b is equal to 2,
we can go back and find a from the first equation.
The next step is to substitute 2 for b,
which gives us 2 times e to the power of -2
plus 4a equals 2.
Let's isolate 4a
by subtracting 2, e to the power of -2 on both sides
which gives us 4a equals 2 minus 2
e to the power of -2.
Divide both sides by 4,
we'll divide everything by 4.
Simplifying, we have a equals 1/2
minus 1/2 times e to the power of -2.
Writing this using positive x
once we can move the e of the power of -2
down to the denominator,
and also write a in the form of 1/2
minus 1 divided by the product of 2 and e squared.
Either form would be acceptable.
So using these values for a and b,
the piecewise function would then be continuous.
And let's go ahead and check this graphically.
So here I've replaced a and b in the original problem
with the values that we found
and the graph is shown on the right.
Notice how the graph is continuous for all real numbers.
We can sketch the graph without lifting up our pencil
because there are no holes or breaks in the graph
making the function continuous.
I hope you found this helpful.