Walk me through the question in the description or at least get me started.

Question

Let f: R -> R be the function defined by the piece-wise function:

{cos (pi x), if x < – 2

f(x) = { ax + b , if –2 <= x <= 2, and

{ x³ , if x > 2

where a and b are (yet undetermined) real constant. Find the values of a and b so that f is continuous on the entire real line. Use the definitions of sequential notions of continuity and limits.

Definition 1:

A sequence of real numbers is a function a: N -> R. Sequences are often written as lists, and we will generally write a(subscript n) = a(n). The term a(subscript n) is the n-th term in the sequence.

Definition 2:

Let a subscript n be sequence. Say the limit as n approaches infinity of

a (subscript n) is L and write

lim as n -> infinity a (subscript n) = L

If for any epsilon > 0, there is some N so large that if n > N, then

| a (subscript n) - L | < epsilon

Answer 1

The question regarding f(x) is unrelated to the definition of a limit for a sequence,

so I ignore the latter.

The function f(x) has three pieces.

For x < -2, f(x) is completely defined and 1 = lim f(x) as x->-2 from the left.

For x > 2, f(x) is also completely defined and 8 = lim f(x) as x->2 from the right.

For -2 <= x <= 2, f(x) is not completely defined as it has the two parameters a,b.

You are to find a, b so that the function f(x) is continuous, meaning that its values are

the same as its limits.

In particular, continuity requires

f(-2) = 1

f(2) = 8

That gives you the two equations

a×(-2) + b = 1

a×2 + b = 8

You can solve them for a, b.