Find all values of a and b so that the following function is continuous for all x ∈ R.
f(x) =
−3a + 4x^5(b) x ≤ −1
ax − 2b −1 < x < 1
3x^2 − bx + a x ≥ 1
To find the values of a and b for which the function is continuous for all x ∈ R, we need to ensure that the function is continuous at the points of transition, which is when x = -1 and x = 1.
At x = -1, we need the right-hand limit to be equal to the left-hand limit. Therefore, we have:
lim(x→-1^-) f(x) = lim(x→-1^+) f(x)
For the left-hand limit, we evaluate the function when x approaches -1 from the left side (x < -1):
lim(x→-1^-) f(x) = lim(x→-1^-) (-3a + 4x^5(b)) = -3a + 4(-1)^5(b) = -3a + 4b
For the right-hand limit, we evaluate the function when x approaches -1 from the right side (-1 < x < 1):
lim(x→-1^+) f(x) = lim(x→-1^+) (ax - 2b) = a(-1) - 2b = -a - 2b
Setting the left-hand limit equal to the right-hand limit:
-3a + 4b = -a - 2b
Simplifying the equation:
2a + 6b = 0 (Equation 1)
Now, let's find the values of a and b for which the function is continuous at x = 1:
lim(x→1^-) f(x) = lim(x→1^+) f(x)
For the left-hand limit, we evaluate the function when x approaches 1 from the left side (-1 < x < 1):
lim(x→1^-) f(x) = lim(x→1^-) (ax - 2b) = a(1) - 2b = a - 2b
For the right-hand limit, we evaluate the function when x approaches 1 from the right side (x > 1):
lim(x→1^+) f(x) = lim(x→1^+) (3x^2 - bx + a) = 3(1)^2 - b(1) + a = 3 - b + a
Setting the left-hand limit equal to the right-hand limit:
a - 2b = 3 - b + a
Simplifying the equation:
b = 3 (Equation 2)
Substituting the value of b from Equation 2 into Equation 1:
2a + 6(3) = 0
2a + 18 = 0
2a = -18
a = -9
Therefore, the values of a and b for which the function is continuous for all x ∈ R are a = -9 and b = 3.