Math / Calculus

QuestionFor $f(x)=x-\ln x$ , and $0.1 \leq x \leq 2$ , find the following. (a) Find the values of $x$ for which $f(x)$ has a local maximum.
Enter your answers in the increasing order. $\begin{array}{l} x=i \\ x=i \end{array}$ eTextbook and Media (b) Find the value of $x$ for which $f(x)$ has a local minimum. $x=\mathbf{i}$ eTextbook and Media (c) Find the value of $x$ for which $f(x)$ has a global maximum. $x=$ $\square$

Studdy Solution

STEP 1

1. The function $f(x) = x - \ln x$ is defined for $x > 0$ since the natural logarithm $\ln x$ is undefined for $x \leq 0$ .
2. We are interested in the interval $0.1 \leq x \leq 2$ .
3. To find local extrema, we need to find the critical points by setting the derivative to zero and analyzing the second derivative.
4. The global maximum will be determined by evaluating the function at critical points and endpoints of the interval.

STEP 2

1. Find the derivative $f'(x)$ .
2. Determine the critical points by solving $f'(x) = 0$ .
3. Use the second derivative test to classify the critical points as local maxima or minima.
4. Evaluate the function at critical points and endpoints to find the global maximum.

STEP 3

Find the derivative of $f(x) = x - \ln x$ .
$f'(x) = 1 - \frac{1}{x}$

STEP 4

Determine the critical points by solving $f'(x) = 0$ .
$1 - \frac{1}{x} = 0$ $1 = \frac{1}{x}$ $x = 1$
The critical point is $x = 1$ .

STEP 5

Use the second derivative test to classify the critical point.
First, find the second derivative:
$f''(x) = \frac{d}{dx}\left(1 - \frac{1}{x}\right) = \frac{1}{x^2}$
Evaluate $f''(x)$ at the critical point $x = 1$ :
$f''(1) = \frac{1}{1^2} = 1$
Since $f''(1) > 0$ , the function has a local minimum at $x = 1$ .

STEP 6

Evaluate the function at the critical point and endpoints to find the global maximum.
Evaluate $f(x)$ at the endpoints $x = 0.1$ and $x = 2$ , and at the critical point $x = 1$ :
$f(0.1) = 0.1 - \ln(0.1) = 0.1 + \ln(10)$ $f(1) = 1 - \ln(1) = 1$ $f(2) = 2 - \ln(2)$
Compare these values to determine the global maximum:
- $f(0.1) = 0.1 + \ln(10) \approx 0.1 + 2.302 \approx 2.402$ - $f(1) = 1$ - $f(2) = 2 - \ln(2) \approx 2 - 0.693 \approx 1.307$
The global maximum occurs at $x = 0.1$ .
(a) Local maximum: None within the interval. (b) Local minimum: $x = 1$ (c) Global maximum: $x = 0.1$

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