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Understanding when a function is continuous at a point
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A function is continuous if you can draw it without lifting your pencil. Mathematically, it's more precise!
A function is continuous at x = a if:
The graph has no breaks, jumps, or holes at that point
A function f is continuous at x = a if all three conditions hold:
Determine if is continuous at x = 3.
| Section | Format | Questions | Time | Weight | Calculator |
|---|---|---|---|---|---|
| Multiple Choice (No Calculator) | MCQ | 30 | 60 min | 33.3% | ๐ซ |
| Multiple Choice (Calculator) | MCQ | 15 | 45 min | 16.7% | โ |
| Free Response (Calculator) | FRQ | 2 | 30 min | 16.7% | โ |
| Free Response (No Calculator) | FRQ | 4 | 60 min | 33.3% | ๐ซ |
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A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of cm/s. How fast is the area of the circle increasing when the radius is cm?
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All three must be true! If even one fails, the function is discontinuous at that point.
Example 1: Is continuous at x = 3?
Check 1: Does f(3) exist? โ Yes
Check 2: Does exist? โ Yes
Check 3: Does ? โ Yes
Conclusion: f is continuous at x = 3!
Is this continuous at x = 2?
Check 1: Does f(2) exist? โ Yes
Check 2: Does exist? โ Yes
Check 3: Does ? โ NO!
Conclusion: f is NOT continuous at x = 2 (even though f(2) exists and the limit exists!)
Many familiar functions are continuous everywhere:
Functions are typically discontinuous where:
Continuous functions have nice properties:
On a graph, a function is continuous at a point if:
Check 1: Does f(3) exist?
This is undefined! โ
Conclusion: f is NOT continuous at x = 3 because f(3) doesn't exist.
The function has a hole at x = 3.
Note: Even though we can find the limit:
The limit exists (equals 6), but f(3) doesn't exist, so the function fails condition 1 of continuity.
For what value of k is continuous at x = 2?
For f to be continuous at x = 2, we need:
Find f(2): Since , we use the second piece:
Is f(x) = xยฒ - 3x + 5 continuous at x = 2?
Step 1: Check the three conditions for continuity at x = a:
Step 2: Check if f(2) is defined: f(2) = (2)ยฒ - 3(2) + 5 = 4 - 6 + 5 = 3 โ f(2) is defined
Step 3: Find lim(xโ2) f(x): Since this is a polynomial, use direct substitution lim(xโ2) (xยฒ - 3x + 5) = 3 โ Limit exists
Step 4: Compare: lim(xโ2) f(x) = 3 = f(2) โ All three conditions satisfied
Answer: Yes, f(x) is continuous at x = 2
Determine if f(x) = { x + 1, if x < 3; 5, if x = 3; 2x - 1, if x > 3 } is continuous at x = 3.
Step 1: Check if f(3) is defined: f(3) = 5 โ
Step 2: Find lim(xโ3โป) f(x): For x < 3, use f(x) = x + 1 lim(xโ3โป) (x + 1) = 3 + 1 = 4
Step 3: Find lim(xโ3โบ) f(x): For x > 3, use f(x) = 2x - 1 lim(xโ3โบ) (2x - 1) = 2(3) - 1 = 5
Step 4: Check if two-sided limit exists: lim(xโ3โป) f(x) = 4 lim(xโ3โบ) f(x) = 5 Since 4 โ 5, lim(xโ3) f(x) does not exist โ
Step 5: Conclusion: Since the limit doesn't exist, f is not continuous at x = 3
Answer: No, f(x) is not continuous at x = 3
Find the value of k that makes f(x) = { (xยฒ - 4)/(x - 2), if x โ 2; k, if x = 2 } continuous at x = 2.
Step 1: For continuity at x = 2, need: lim(xโ2) f(x) = f(2) = k
Step 2: Find lim(xโ2) (xยฒ - 4)/(x - 2): Direct substitution gives 0/0 (indeterminate)
Step 3: Factor numerator: xยฒ - 4 = (x - 2)(x + 2)
Step 4: Simplify: (xยฒ - 4)/(x - 2) = [(x - 2)(x + 2)]/(x - 2) = x + 2 (for x โ 2)
Step 5: Evaluate limit: lim(xโ2) (x + 2) = 2 + 2 = 4
Step 6: Set equal to f(2): k = 4
Step 7: Verify: With k = 4, lim(xโ2) f(x) = 4 = f(2) โ
Answer: k = 4
Find the limit from the left:
Find the limit from the right:
For continuity, these must all be equal:
Answer:
With this value, f(2) = 4 and both one-sided limits equal 4, making the function continuous!