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The function f(x)=tanx+(1)/(x), AA x in...

The function `f(x)=tanx+(1)/(x), AA x in (0, (pi)/(2))` has

A

one local maximum

B

one local minimum

C

one local maximum and one minimum

D

no local maximum of minimum

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the function \( f(x) = \tan x + \frac{1}{x} \) for \( x \) in the interval \( (0, \frac{\pi}{2}) \). ### Step 1: Find the derivative of the function We start by finding the derivative \( f'(x) \) of the function \( f(x) \). \[ f'(x) = \frac{d}{dx}(\tan x) + \frac{d}{dx}\left(\frac{1}{x}\right) \] Using the derivatives of \( \tan x \) and \( \frac{1}{x} \): \[ f'(x) = \sec^2 x - \frac{1}{x^2} \] ### Step 2: Set the derivative to zero To find critical points, we set the derivative equal to zero: \[ \sec^2 x - \frac{1}{x^2} = 0 \] This can be rearranged to: \[ \sec^2 x = \frac{1}{x^2} \] ### Step 3: Rewrite in terms of sine and cosine Recall that \( \sec x = \frac{1}{\cos x} \), so we can rewrite \( \sec^2 x \): \[ \frac{1}{\cos^2 x} = \frac{1}{x^2} \] Cross-multiplying gives: \[ x^2 = \cos^2 x \] ### Step 4: Analyze the function We need to analyze the behavior of \( f'(x) \) in the interval \( (0, \frac{\pi}{2}) \). 1. As \( x \to 0 \), \( \tan x \to 0 \) and \( \frac{1}{x} \to \infty \), thus \( f(x) \to \infty \). 2. As \( x \to \frac{\pi}{2} \), \( \tan x \to \infty \) and \( \frac{1}{x} \to \frac{2}{\pi} \), thus \( f(x) \to \infty \). ### Step 5: Determine the nature of critical points To determine whether the critical point found is a minimum or maximum, we can analyze the sign of \( f'(x) \): - For \( x \) close to 0, \( \sec^2 x \) is large and positive, while \( -\frac{1}{x^2} \) is negative. Thus, \( f'(x) > 0 \). - For \( x \) approaching \( \frac{\pi}{2} \), \( \sec^2 x \) becomes very large, and \( -\frac{1}{x^2} \) becomes less significant. Thus, \( f'(x) > 0 \). Since \( f'(x) \) changes from positive to negative, there is a local minimum in the interval. ### Conclusion The function \( f(x) = \tan x + \frac{1}{x} \) has **one local minimum** in the interval \( (0, \frac{\pi}{2}) \).
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