1.3 continuity, end behavior, and limits. ex. 1 determine whether each function is continuous at...
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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 - ∞