when x = -1
4 - 2 x - x^2 = a x + b
4 +2 -1 = -a + b
5 = -a + b
then when x= +1
a x + b = x^3
a + b = 1
so
two equations
a + b = 1
-a + b = 5
------------------- add them
2 b = 6
b =3
a = -2
4 - 2 x - x^2 = a x + b
4 +2 -1 = -a + b
5 = -a + b
then when x= +1
a x + b = x^3
a + b = 1
so
two equations
a + b = 1
-a + b = 5
------------------- add them
2 b = 6
b =3
a = -2
At x = -1, we have f(x) = 4 - 2(-1) - (-1)^2, which simplifies to f(-1) = 7.
Substituting x = -1 into the middle piece of the function, we get -a + b = 7.
At x = 1, we have f(x) = 1^3, which simplifies to f(1) = 1.
Substituting x = 1 into the middle piece of the function, we get a + b = 1.
Now we have a system of equations:
-a + b = 7
a + b = 1
Adding these two equations together, we get 2b = 8, which means b = 4.
Substituting this value of b into the second equation, we get a + 4 = 1, so a = -3.
Therefore, the values for a and b that make the piecewise function continuous are a = -3 and b = 4.
First, let's find the value of f(x) for x = -1. We plug in x = -1 into the first equation:
f(-1) = 4 - 2(-1) - (-1)^2
= 4 + 2 - 1
= 5
Now, let's find the value of f(x) for x = 1. We plug in x = 1 into the second equation:
f(1) = a(1) + b
= a + b
Since f(x) on the left and right intervals should match at x = -1 and x = 1, we set the values equal to each other.
5 = a + b
Now, we can substitute the value of f(1) into the equation to solve for a:
5 = a + b
Since we don't have enough information to solve for a and b, we cannot find unique values for the constants a and b that make the piecewise function continuous for all real numbers.
First, let's find the value of the function at x = -1 by substituting it into the first part of the piecewise function:
f(-1) = 4 - 2(-1) - (-1)^2
= 4 + 2 - 1
= 5
Now, let's find the value of the function at x = 1 by substituting it into the second part of the piecewise function:
f(1) = a(1) + b
= a + b
Since the function needs to be continuous at x = -1 and x = 1, the values of the function at these points should be the same. Therefore, we have the following equations:
5 = a + b (at x = -1)
a + b = 1 (at x = 1)
To solve this system of equations, we can substitute the value of b from the first equation into the second equation:
a + (5 - a) = 1
5 - a + a = 1
5 = 1
This equation is not solvable, which means that there are no values of a and b that make the piecewise function continuous for all real numbers.