Math Problem Statement
Solution
We are tasked with finding the values of and that make the function continuous on the whole real line. The given piecewise function is:
\begin{cases} \frac{2}{\pi} \arccos\left(e^{\frac{1}{x+3}}\right), & x < -3 \\ e^{ax + b}, & -3 \leq x \leq 2 \\ \frac{x-2}{e^{\sqrt{23 + x^2} - 5}}, & x > 2 \end{cases}$$ ### Steps to ensure continuity: 1. **Continuity at $$x = -3$$:** For $$f(x)$$ to be continuous at $$x = -3$$, the left-hand limit as $$x \to -3^-$$ from the first part of the function must equal the right-hand limit as $$x \to -3^+$$ from the second part of the function. So we equate: $$\lim_{x \to -3^-} \frac{2}{\pi} \arccos\left(e^{\frac{1}{x+3}}\right) = \lim_{x \to -3^+} e^{ax + b}$$ As $$x \to -3^-$$, $$e^{\frac{1}{x+3}}$$ tends to $$e^{-\infty}$$, which is 0. Therefore, we have: $$\frac{2}{\pi} \arccos(0) = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1$$ So, the left-hand limit at $$x = -3$$ is 1. Now, the right-hand limit from the second piece as $$x \to -3^+$$ is: $$e^{a(-3) + b} = e^{-3a + b}$$ For continuity at $$x = -3$$, these limits must be equal: $$e^{-3a + b} = 1$$ Taking the natural logarithm of both sides: $$-3a + b = 0 \quad \Rightarrow \quad b = 3a$$ 2. **Continuity at $$x = 2$$:** For continuity at $$x = 2$$, the second and third parts of the function must be equal at $$x = 2$$. That is: $$\lim_{x \to 2^-} e^{ax + b} = \lim_{x \to 2^+} \frac{x - 2}{e^{\sqrt{23 + x^2} - 5}}$$ Evaluating the third piece at $$x = 2$$: $$\frac{2 - 2}{e^{\sqrt{23 + 4} - 5}} = \frac{0}{e^{\sqrt{27} - 5}} = 0$$ Therefore, the right-hand limit as $$x \to 2^+$$ is 0. For continuity at $$x = 2$$, the second piece must also be 0 at $$x = 2$$: $$e^{2a + b} = 0$$ Since the exponential function can never be zero, this implies that the only solution is $$2a + b = -\infty$$, which is not possible. Therefore, the function is continuous only if $$e^{2a + b} = 0$$, which suggests we seek other possible limits for special values. ThusAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Piecewise Functions
Limits
Exponential Functions
Arccos Function
Formulas
f(x) = (2/π) arccos(e^(1/(x+3))) for x < -3
f(x) = e^(ax + b) for -3 ≤ x ≤ 2
f(x) = (x - 2)/(e^(√(23 + x^2) - 5)) for x > 2
Limit comparison at x = -3 and x = 2 for continuity
Theorems
Limit definition of continuity
Natural logarithm properties
Suitable Grade Level
Undergraduate (Calculus or Advanced Algebra)
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