Math Problem Statement
Which of the following functions are continuous on (-∞, ∞)? (i) f(x) = {x+2, x ≤ 1; 3 + 5 ln x, x > 1}, (ii) f(x) = {7e^x - 4, x < 0; 3 cos x, x ≥ 0}, (iii) f(x) = {(x^2 - 49)/(x - 7), x ≠ 7; 12, x = 7}
Solution
Let's analyze each function to determine if they are continuous.
1. Function (i):
\begin{cases} x + 2 & \text{for } x \leq 1 \\ 3 + 5 \ln(x) & \text{for } x > 1 \end{cases}$$ - For $$x \leq 1$$, $$f(x) = x + 2$$ is a linear function, and hence continuous everywhere in this domain. - For $$x > 1$$, $$f(x) = 3 + 5 \ln(x)$$ is continuous where $$x > 1$$, since the logarithm function is continuous for $$x > 0$$. - To check the continuity at $$x = 1$$, we need to confirm that: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$ - For $$x \to 1^-$$, $$f(x) = x + 2$$, so $$\lim_{x \to 1^-} f(x) = 1 + 2 = 3$$. - For $$x \to 1^+$$, $$f(x) = 3 + 5 \ln(x)$$, so $$\lim_{x \to 1^+} f(x) = 3 + 5 \ln(1) = 3$$. - Thus, $$f(x)$$ is continuous at $$x = 1$$. So, function (i) is continuous for all $$x$$. ### 2. Function (ii): $$f(x) = \begin{cases} 7e^x - 4 & \text{for } x < 0 \\ 3 \cos(x) & \text{for } x \geq 0 \end{cases}$$ - For $$x < 0$$, $$f(x) = 7e^x - 4$$ is an exponential function, which is continuous everywhere. - For $$x \geq 0$$, $$f(x) = 3 \cos(x)$$ is a cosine function, which is also continuous everywhere. - To check the continuity at $$x = 0$$, we need to confirm that: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$$ - For $$x \to 0^-$$, $$\lim_{x \to 0^-} f(x) = 7e^0 - 4 = 7 - 4 = 3$$. - For $$x \to 0^+$$, $$f(x) = 3 \cos(x)$$, so $$\lim_{x \to 0^+} f(x) = 3 \cos(0) = 3$$. - Thus, $$f(x)$$ is continuous at $$x = 0$$. So, function (ii) is continuous for all $$x$$. ### 3. Function (iii): $$f(x) = \begin{cases} \frac{x^2 - 49}{x - 7} & \text{for } x \neq 7 \\ 12 & \text{for } x = 7 \end{cases}$$ - For $$x \neq 7$$, $$f(x) = \frac{x^2 - 49}{x - 7}$$ simplifies to $$f(x) = x + 7$$, which is continuous everywhere for $$x \neq 7$$. - To check the continuity at $$x = 7$$, we need to confirm that: $$\lim_{x \to 7} f(x) = f(7)$$ - $$\lim_{x \to 7} f(x) = \lim_{x \to 7} (x + 7) = 7 + 7 = 14$$. - But $$f(7) = 12$$, so $$f(x)$$ is **not continuous** at $$x = 7$$. Thus, function (iii) is not continuous at $$x = 7$$. ### Conclusion: The functions that are continuous are: - (i) Continuous everywhere - (ii) Continuous everywhere - (iii) Not continuous at $$x = 7$$ Thus, the correct answer is **E) (ii) and (iii) only**. Would you like further details on any of the steps or functions? --- Here are some related questions: 1. What does continuity mean for a function? 2. How do you check the continuity at a specific point? 3. What is the difference between a piecewise and a non-piecewise function regarding continuity? 4. How do exponential and logarithmic functions behave in terms of continuity? 5. Can a function be continuous at all points except one? **Tip**: Always check the limits from both sides at a point where the definition of the function changes.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Piecewise Functions
Limits
Formulas
f(x) = x + 2 for x ≤ 1, f(x) = 3 + 5 ln(x) for x > 1
f(x) = 7e^x - 4 for x < 0, f(x) = 3 cos(x) for x ≥ 0
f(x) = (x^2 - 49)/(x - 7) for x ≠ 7, f(x) = 12 for x = 7
Theorems
Continuity at a point: The limit from the left and the right must equal the function's value
Suitable Grade Level
Grades 11-12
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