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Find the critical numbers, the intervals on which \( f(x) \) is incie \( f(x)=x+\frac{9}{x} \) Select the correct choice below and, if necessary, fill in the A. The critical number(s) is(are) E. The function has no critical numbers. Select the correct choice below and, if necessary, fill in the A. The function is increasing on \( \square \). B. The function is never increasing. B. Type your answer in interval notation. Type integers B. Thernma

Ask by Barnett Summers. in the United States
Mar 10,2025

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Answer

The critical numbers are -3 and 3. The function is increasing on the intervals (-∞, -3) and (3, ∞).

Solution

**Critical Numbers** 1. The derivative of the function is \[ f'(x)=1-\frac{9}{x^2}. \] 2. Set the derivative equal to zero to find the critical numbers: \[ 1-\frac{9}{x^2}=0 \quad \Rightarrow \quad \frac{9}{x^2}=1 \quad \Rightarrow \quad x^2=9 \quad \Rightarrow \quad x=3 \text{ or } x=-3. \] Therefore, the critical numbers are \( -3 \) and \( 3 \). **Intervals of Increase** 1. To determine where the function is increasing, solve the inequality: \[ f'(x)>0 \quad \Rightarrow \quad 1-\frac{9}{x^2}>0. \] 2. Rearranging the inequality gives: \[ 1 > \frac{9}{x^2} \quad \Rightarrow \quad x^2 > 9. \] 3. Taking the square root of both sides, we have: \[ |x| > 3, \] which means \[ x<-3 \quad \text{or} \quad x>3. \] 4. Therefore, the function \( f(x) \) is increasing on the intervals: \[ (-\infty,-3) \quad \text{and} \quad (3,\infty). \] **Final Answers:** - The critical numbers are \( -3 \) and \( 3 \). - The function is increasing on the intervals \( (-\infty,-3) \) and \( (3,\infty) \).

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To find the critical numbers of the function \( f(x) = x + \frac{9}{x} \), we first need to calculate the derivative and find where it is equal to zero or undefined. The derivative \( f'(x) = 1 - \frac{9}{x^2} \). Setting this equal to zero gives \( 1 - \frac{9}{x^2} = 0 \), leading to \( x^2 = 9 \) so \( x = 3 \) or \( x = -3 \). However, since \( x \) must not equal zero in the original function, our critical numbers are \( 3 \) and \( -3 \). Next, to determine where the function is increasing, we analyze the sign of the derivative \( f'(x) \). The intervals can be tested using test points: for \( x < -3 \), \( f'(x) > 0 \) (increasing); for \( -3 < x < 0 \), \( f'(x) < 0 \) (decreasing); and for \( x > 0 \) (including \( 3 \)), \( f'(x) > 0 \) (increasing). Therefore, the function increases on the intervals \( (-\infty, -3) \) and \( (3, \infty) \). During the 19th century, mathematicians began formalizing the concept of derivatives, marking a significant shift in calculus. This period saw the groundwork for understanding functions and their behavior, laying down theories still used in calculus today, emphasized by figures such as Newton and Leibniz. Applications of derivatives are found throughout modern life—from computing the speed of a car (instantaneous rate of change) to optimizing various processes in industries. Engineers rely on calculus to maximize efficiency and design more effective systems, showing the powerful real-world utility of understanding critical points and intervals of increase and decrease in functions.

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