Pregunta
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Find the critical numbers, the intervals on which \( f(x) \) is increasing, the \( f(x)=\frac{x^{2}}{x-5} \) Select the correct choice below and, if necessary, fill in the answer b A. The critical number(s) of the function are \( x=\square \). (Use a comma lo separate answers as necded.) B. There is no solution. Select the correct choice below and, if necessary, fill in the answer b A. The function is increasing on \( \square \). B. There is no solution.

Ask by Peters Campbell. in the United States
Mar 10,2025

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A. The critical numbers are \(0\) and \(10\). B. The function is increasing on \((-\infty, 0)\) and \((10, \infty)\).

Solución

We begin with the function \[ f(x)=\frac{x^2}{x-5}. \] Note that the domain is all real numbers except \(x=5\) (since the denominator cannot be zero). **Step 1. Compute the derivative \(f'(x)\).** Using the quotient rule, if \[ f(x)=\frac{u(x)}{v(x)}, \] then \[ f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}. \] Here, \[ u(x)=x^2, \quad u'(x)=2x, \quad v(x)=x-5, \quad v'(x)=1. \] So, \[ f'(x)=\frac{2x(x-5)-x^2\cdot 1}{(x-5)^2}=\frac{2x(x-5)-x^2}{(x-5)^2}. \] Simplify the numerator: \[ 2x(x-5)-x^2=2x^2-10x-x^2=x^2-10x. \] Thus, \[ f'(x)=\frac{x^2-10x}{(x-5)^2}=\frac{x(x-10)}{(x-5)^2}. \] **Step 2. Find the critical numbers.** Critical numbers occur when \(f'(x)=0\) or when \(f'(x)\) is undefined (provided the point is in the domain). 1. \(f'(x)=0\) when the numerator is zero: \[ x(x-10)=0 \quad \Longrightarrow \quad x=0 \quad \text{or} \quad x=10. \] 2. \(f'(x)\) is undefined when the denominator is zero: \[ (x-5)^2=0 \quad \Longrightarrow \quad x=5. \] However, \(x=5\) is not in the domain of \(f\). Thus, the critical numbers are: \[ x=0 \quad \text{and} \quad x=10. \] **Step 3. Determine where \(f(x)\) is increasing.** The function is increasing where \(f'(x) > 0\). Since the denominator \((x-5)^2\) is always positive (except at \(x=5\) where \(f\) is not defined), the sign of \(f'(x)\) is determined by the numerator \(x(x-10)\). Solve the inequality: \[ x(x-10)>0. \] This product is positive when: - Both factors are positive: \(x>0\) and \(x-10>0\) which gives \(x>10\). - Both factors are negative: \(x<0\) and \(x-10<0\) which gives \(x<0\). Thus, \(f(x)\) is increasing on: \[ (-\infty,0) \quad \text{and} \quad (10,\infty). \] **Final Answers** A. The critical number(s) of the function are \(0, 10\). B. The function is increasing on \((-\infty, 0)\) and \((10, \infty)\).

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To find the critical numbers of the function \( f(x)=\frac{x^{2}}{x-5} \), we first need to find the derivative \( f'(x) \) and set it equal to zero. Calculating the derivative using the quotient rule yields: \[ f'(x) = \frac{(x-5)(2x) - x^2(1)}{(x-5)^2} = \frac{2x^2 - 10x - x^2}{(x-5)^2} = \frac{x^2 - 10x}{(x-5)^2} \] Setting the numerator equal to zero gives \( x^2 - 10x = 0 \) which factors to \( x(x - 10) = 0 \). Thus, the critical numbers are \( x = 0 \) and \( x = 10 \). Now, let's assess the behavior of \( f(x) \) around these points to find the intervals of increase. Test the intervals: 1. Choose \( x < 0 \) (e.g., \( x = -1 \)), \( f'(-1) > 0 \) (increasing). 2. Choose \( 0 < x < 5 \) (e.g., \( x = 1 \)), \( f'(1) < 0 \) (decreasing). 3. Choose \( 5 < x < 10 \) (e.g., \( x = 6 \)), \( f'(6) > 0 \) (increasing). 4. Choose \( x > 10 \) (e.g., \( x = 11 \)), \( f'(11) < 0 \) (decreasing). Thus, the function \( f(x) \) is increasing on the intervals \( (-\infty, 0) \) and \( (5, 10) \). A. The critical number(s) of the function are \( x=0, 10 \). A. The function is increasing on \( (-\infty, 0) \) and \( (5, 10) \).

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