Math  /  Calculus

QuestionIn Problems 9-16, use the graph of the function ff she mate the indicated limits and function values.
A 9. f(0.5)f(-0.5) Figure for 9169-16
11. f(1.75)f(1.75)
10. f(0.5)f(0.5)
13. (A) limx0f(x)\lim _{x \rightarrow 0^{-}} f(x)
12. f(2.25)f(2.25) (C) limx0f(x)\lim _{x \rightarrow 0} f(x) (B) limx0+f(x)\lim _{x \rightarrow 0^{+}} f(x)
14. (A) limx1f(x)\lim _{x \rightarrow 1^{-}} f(x) (D) f(0)f(0) (C) limx1f(x)\lim _{x \rightarrow 1} f(x) (B) limx1+f(x)\lim _{x \rightarrow 1^{+}} f(x)
15. (A) limx2f(x)\lim _{x \rightarrow 2^{-}} f(x) (D) f(1)f(1) (C) limx2f(x)\lim _{x \rightarrow 2} f(x) (B) limx2+f(x)\lim _{x \rightarrow 2^{+}} f(x)
16. (A) limx4f(x)\lim _{x \rightarrow 4^{-}} f(x) (D) f(2)f(2) (C) limx4f(x)\lim _{x \rightarrow 4} f(x) (B) limx4+f(x)\lim _{x \rightarrow 4^{+}} f(x)

In Problems 17-21 (D) f(4)f(4)

Studdy Solution
Evaluate f(4) f(4) .
From the graph, at x=4 x = 4 , there is no visible discontinuity, so the function value is equal to the limit. Therefore, f(4)=4 f(4) = 4 .
Solution Summary:
1. f(0.5)=1 f(-0.5) = 1
2. f(0.5)=0.5 f(0.5) = 0.5
3. f(1.75)=0.75 f(1.75) = -0.75
4. f(2.25)=2.25 f(2.25) = 2.25
5. limx0f(x)=1 \lim_{x \to 0^-} f(x) = 1
6. limx0+f(x)=1 \lim_{x \to 0^+} f(x) = 1
7. limx0f(x)=1 \lim_{x \to 0} f(x) = 1
8. f(0) does not exist f(0) \text{ does not exist}
9. limx1f(x)=0 \lim_{x \to 1^-} f(x) = 0
10. limx1+f(x)=0 \lim_{x \to 1^+} f(x) = 0
11. limx1f(x)=0 \lim_{x \to 1} f(x) = 0
12. f(1)=0 f(1) = 0
13. limx2f(x)=1 \lim_{x \to 2^-} f(x) = -1
14. limx2+f(x)=2 \lim_{x \to 2^+} f(x) = 2
15. limx2f(x) does not exist \lim_{x \to 2} f(x) \text{ does not exist}
16. f(2)=2 f(2) = 2
17. limx4f(x)=4 \lim_{x \to 4^-} f(x) = 4
18. limx4+f(x)=4 \lim_{x \to 4^+} f(x) = 4
19. limx4f(x)=4 \lim_{x \to 4} f(x) = 4
20. f(4)=4 f(4) = 4

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