1.3 Continuity, End Behavior, and Limits. Ex. 1 Determine whether each function is continuous at the given x value(s). Justify using the continuity.

1.3 – Continuity, End Behavior, and Limits

Ex. 1 Determine whether each function is continuous at the given x value(s). Justify using the continuity test. If discontinuous, identify the type of discontinuity as infinite, jump, or removable.

a. f(x) = 3x – 2 if x > -3 ; at x = -3 2 – x if x < - 3

Ex. 1 Determine whether each function is continuous at the given x value(s). Justify using the continuity test. If discontinuous, identify the type of discontinuity as infinite, jump, or removable.

a. f(x) = 3x – 2 if x > -3 ; at x = -3 2 – x if x < - 3

1. Find f(-3).

Ex. 1 Determine whether each function is continuous at the given x value(s). Justify using the continuity test. If discontinuous, identify the type of discontinuity as infinite, jump, or removable.

a. f(x) = 3x – 2 if x > -3 ; at x = -3 2 – x if x < - 3

1. Find f(-3). f(-3) = 2 – (-3) = 5

Ex. 1 Determine whether each function is continuous at the given x value(s). Justify using the continuity test. If discontinuous, identify the type of discontinuity as infinite, jump, or removable.

a. f(x) = 3x – 2 if x > -3 ; at x = -3 2 – x if x < - 3

1. Find f(-3). f(-3) = 2 – (-3) = 5, so f(-3) exists

2. Investigate values close to f(-3)

2. Investigate values close to f(-3)

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) 5.1 5.01 5.001 -10.997 -10.97 -10.7

2. Investigate values close to f(-3)

As x -3 from left, f(x) 5

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) 5.1 5.01 5.001 -10.997 -10.97 -10.7

2. Investigate values close to f(-3)

As x -3 from left, f(x) 5As x -3 from right, f(x) -11

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) 5.1 5.01 5.001 -10.997 -10.97 -10.7

2. Investigate values close to f(-3)

As x -3 from left, f(x) 5As x -3 from right, f(x) -11Since don’t approach same value,

discontinuous and jump discontinuity.

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) 5.1 5.01 5.001 -10.997 -10.97 -10.7

b. f(x) = x + 3 ; at x = -3 and x = 3 x2 – 9

b. f(x) = x + 3 ; at x = -3 and x = 3 x2 – 9

1. Find f(-3) and f(3).

b. f(x) = x + 3 ; at x = -3 and x = 3 x2 – 9

1. Find f(-3) and f(3).f(-3) = -3 + 3 = 0 = Ø

(-3)2 – 9 0

b. f(x) = x + 3 ; at x = -3 and x = 3 x2 – 9

1. Find f(-3) and f(3).f(-3) = -3 + 3 = 0 = Ø

(-3)2 – 9 0f(3) = 3 + 3 = 6 = Ø (3)2 – 9 0

b. f(x) = x + 3 ; at x = -3 and x = 3 x2 – 9

1. Find f(-3) and f(3).f(-3) = -3 + 3 = 0 = Ø

(-3)2 – 9 0f(3) = 3 + 3 = 6 = Ø (3)2 – 9 0

Since both f(-3) = Ø and f(3) = Ø, f(x) is discontinuous at both x = -3 and x = 3.

2. Investigate values close to f(-3) and f(3).

2. Investigate values close to f(-3) and f(3).

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) -0.164 -0.166 -0.167 -0.167 -0.167 -0.169

2. Investigate values close to f(-3) and f(3).

As x -3 from left, f(x) -0.167

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) -0.164 -0.166 -0.167 -0.167 -0.167 -0.169

2. Investigate values close to f(-3) and f(3).

As x -3 from left, f(x) -0.167As x -3 from right, f(x) -0.167

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) -0.164 -0.166 -0.167 -0.167 -0.167 -0.169

2. Investigate values close to f(-3) and f(3).

As x -3 from left, f(x) -0.167As x -3 from right, f(x) -0.167

Since they approach same value, limit exists.

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) -0.164 -0.166 -0.167 -0.167 -0.167 -0.169

2. Investigate values close to f(-3) and f(3).

As x -3 from left, f(x) -0.167As x -3 from right, f(x) -0.167

Since they approach same value, limit exists.

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) -0.164 -0.166 -0.167 -0.167 -0.167 -0.169

x 2.9 2.99 2.999 3.0 3.001 3.01 3.1

f(x) -10 -100 -1000 1000 100 10

2. Investigate values close to f(-3) and f(3).

As x -3 from left, f(x) -0.167As x -3 from right, f(x) -0.167

Since they approach same value, limit exists.

As x 3 from left, f(x) -∞

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) -0.164 -0.166 -0.167 -0.167 -0.167 -0.169

x 2.9 2.99 2.999 3.0 3.001 3.01 3.1

f(x) -10 -100 -1000 1000 100 10

2. Investigate values close to f(-3) and f(3).

As x -3 from left, f(x) -0.167As x -3 from right, f(x) -0.167

Since they approach same value, limit exists.

As x 3 from left, f(x) -∞As x 3 from right, f(x) ∞

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) -0.164 -0.166 -0.167 -0.167 -0.167 -0.169

x 2.9 2.99 2.999 3.0 3.001 3.01 3.1

f(x) -10 -100 -1000 1000 100 10

2. Investigate values close to f(-3) and f(3).

As x -3 from left, f(x) -0.167As x -3 from right, f(x) -0.167

Since they approach same value, limit exists.

As x 3 from left, f(x) -∞As x 3 from right, f(x) ∞

Since limit x -3 exists but f(-3) doesn’t, removable discontinuity.

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) -0.164 -0.166 -0.167 -0.167 -0.167 -0.169

x 2.9 2.99 2.999 3.0 3.001 3.01 3.1

f(x) -10 -100 -1000 1000 100 10

2. Investigate values close to f(-3) and f(3).

As x -3 from left, f(x) -0.167As x -3 from right, f(x) -0.167

Since they approach same value, limit exists.

As x 3 from left, f(x) -∞As x 3 from right, f(x) ∞

Since limit x -3 exists but f(-3) doesn’t, removable discontinuity.Since limit x -3 doesn’t exist, infinite discontinuity.

x -3.1 -3.01 -3.001 -3 -2.999 -2.99 -2.9

f(x) -0.164 -0.166 -0.167 -0.167 -0.167 -0.169

x 2.9 2.99 2.999 3.0 3.001 3.01 3.1

f(x) -10 -100 -1000 1000 100 10

Ex. 2 Use the graph of the function to describe its end behavior.

Ex. 2 Use the graph of the function to describe its end behavior.

lim f(x) = - ∞ x - ∞

Ex. 2 Use the graph of the function to describe its end behavior.

lim f(x) = - ∞ x - ∞

lim f(x) = - ∞ x - ∞