Not sure how do do this. Please help?

Find all values of a and b so that the function is continuous for all x E R.

f(x)= −3a + 4(x^5)b x ≤ −1

{ ax − 2b −1 < x < 1

3x^2 − bx + a x ≥ 1

The question did not come out as clear as it was typed. It is:

−3a + 4(x^5)b, x ≤ −1

ax − 2b −1, −1 < x < 1

3x^2 − bx + a, x ≥ 1

for f to be continuous, all three definitions must be equal at x=1. Since f is defined at x=1 in three different ways, we need

-3a+4b = a-2b-1
a-2b-1 = 3-b+a

So just solve those for a and b

To find the values of a and b that would make the function continuous for all x in the real numbers (R), we need to consider the left-hand limit, right-hand limit, and the value of the function at each breakpoint.

1. Left-hand limit (x approaching -1):
For the function to be continuous at x = -1, the left-hand limit should be equal to the value of the function at x = -1. We can find the left-hand limit by plugging in -1 as the value of x into the first piece of the function:

lim(x→-1^-) f(x) = -3a + 4(-1^5)b

2. Right-hand limit (x approaching 1):
Similarly, for the function to be continuous at x = 1, the right-hand limit should be equal to the value of the function at x = 1. We can find the right-hand limit by plugging in 1 as the value of x into the last piece of the function:

lim(x→1^+) f(x) = 3(1^2) - b(1) + a

3. Value of the function at the breakpoint x = -1:
f(-1) = a*(-1) - 2b

4. Value of the function at the breakpoint x = 1:
f(1) = 3(1^2) - b(1) + a

Using these expressions, we can set up a system of equations and solve for a and b.

-3a + 4(-1^5)b = a*(-1) - 2b (Equation 1)
3(1^2) - b(1) + a = 3(1^2) - b(1) + a (Equation 2)

We have two equations with two variables, which we can solve using algebraic techniques such as substitution or elimination.