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We are tasked with finding the values of \(a\) and \(b\) that make the function \(f(x)\) continuous on the whole real line. The given piecewise function is:

\[
f(x) = 
\begin{cases} 
\frac{2}{\pi} \arccos\left(e^{\frac{1}{x+3}}\right), & x < -3 \\
e^{ax + b}, & -3 \leq x \leq 2 \\
\frac{x-2}{e^{\sqrt{23 + x^2} - 5}}, & x > 2
\end{cases}
\]

### 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.

Thus

Find the values of a, b ∈ ℝ that make the function f continuous on the whole real line. The piecewise function f(x) is given as: f(x) = { (2/π) arccos(e^(1/(x+3))), x < -3; e^(ax + b), -3 ≤ x ≤ 2; (x - 2)/(e^(√(23 + x^2) - 5)), x > 2 }.

This problem explores finding the values of a and b that ensure continuity of a piecewise function. The function involves arccos, exponential functions, and limits. Step-by-step explanation for ensuring continuity at specific points (x = -3 and x = 2) is provided.

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