Example Question - continuous function
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Determining Values for Function Continuity
The image shows a piecewise function with three different expressions for different intervals of x: 1. \(f(x) = \log_2(3x + a)\) for \(x < 1\) 2. \(f(x) = 3^x\) for \(x = 1\) 3. \(f(x) = \sqrt{x + b}\) for \(x > 1\) The question asks for the values of \(a\) and \(b\) such that the function \(f(x)\) is continuous at \(x = 1\). To ensure continuity at \(x = 1\), the following conditions must be met: 1. The limit of \(f(x)\) as \(x\) approaches 1 from the left must equal the value of \(f(1)\). 2. The limit of \(f(x)\) as \(x\) approaches 1 from the right must also equal the value of \(f(1)\). Let’s find these limits: 1. The limit as \(x\) approaches 1 from the left: \[ \lim_{{x \to 1^-}} \log_2(3x + a) = \log_2(3 \cdot 1 + a) = \log_2(3 + a) \] 2. The limit as \(x\) approaches 1 from the right: \[ \lim_{{x \to 1^+}} \sqrt{x + b} = \sqrt{1 + b} \] We also know that \(f(1) = 3^1 = 3\). Setting these limits equal to \(f(1)\), we get: \[ \log_2(3 + a) = 3 \] \[ \sqrt{1 + b} = 3 \] From the first equation, we can solve for \(a\): \[ 2^3 = 3 + a \] \[ 8 = 3 + a \] \[ a = 5 \] From the second equation, we can solve for \(b\): \[ 3^2 = 1 + b \] \[ 9 = 1 + b \] \[ b = 8 \] Therefore, the values of \(a\) and \(b\) are 5 and 8, respectively.
Finding Value of 'a' for Continuous Function at x = 2
The function \( f(x) \) is defined by three expressions for different ranges of \( x \), and we are asked to find the value of "a" given that the function is continuous at \( x = 2 \). For a function to be continuous at a point, the left-hand limit (as \( x \) approaches the point from the left), the right-hand limit (as \( x \) approaches the point from the right), and the function's value at the point must all be the same. Let's evaluate the limit from the left (\( x < 2 \)) and from the right (\( x > 2 \)) as well as the function's value at \( x = 2 \): - The right-hand limit as \( x \) approaches 2 from the right is given by the expression \( x + a \). As \( x \) approaches 2, this expression becomes \( 2 + a \). - The function's value at \( x = 2 \) is given as 5. - The left-hand limit as \( x \) approaches 2 from the left is given by the expression \( 2x + 1 \). As \( x \) approaches 2, this expression becomes \( 2(2) + 1 = 4 + 1 = 5 \). Since the function is continuous at \( x = 2 \), the right-hand limit must equal the left-hand limit and also the function's value at \( x = 2 \). So we have the equation \( 2 + a = 5 \). Solving for \( a \): \[ a = 5 - 2 \] \[ a = 3 \] Therefore, the value of "a" that makes the function continuous at \( x = 2 \) is 3.
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