Chapter 3: Q. 82 (page 311)

Determine the local extrema of a function.
$f (x) = x^{2} \ln (0.2 x), I = (0, 4], J = (0, \infty)$

Short Answer

Expert verified

On $I$ , $f$ has a global maximum at $x = 0$ and a global minimum at $x = \frac{5}{\sqrt{e}}$ . On $J$ , $f$ has no global minimum at $x = \frac{5}{\sqrt{e}}$ and it has no global maximum.

Step by step solution

Step 1. Given Information.

The function,

$f (x) = x^{2} \ln (0.2 x), I = (0, 4], J = (0, \infty)$ .

Step 2. The critical point.

For critical points, we consider,

$f' (x) = 0 \frac{d}{d x} (x^{2} \ln (0.2 x)) = 0 x^{2} \times \frac{1}{0.2 x} \times 0.2 + \ln (0.2 x) 2 x = 0 x + 2 x \ln (0.2 x) = 0 x (1 + 2 \ln (0.2 x)) = 0$

We consider,

either $x = 0$

or, $1 + 2 \ln (0.2 x) = 0 2 \ln (0.2 x) = - 1 \ln (0.2 x) = - \frac{1}{2} 0.2 x = e^{- \frac{1}{2}} x = \frac{e^{- \frac{1}{2}}}{0.2} x = \frac{1}{0.2 \sqrt{e}} x = \frac{5}{\sqrt{e}}$

Again,

$\lim_{x \to 0} f (x) = \lim_{x \to 0} (x^{2} \ln (0.2 x)) = \lim_{x \to 0} \frac{\ln (0.2 x)}{\frac{1}{x^{2}}} [\frac{- \infty}{\infty}] = \lim_{x \to 0} \frac{\frac{1}{0.2 x} \times 0.2}{- \frac{2}{x^{3}}} = \lim_{x \to 0} (- \frac{1}{x} \times \frac{x^{3}}{2}) = \lim_{x \to 0} (- \frac{x^{2}}{2})$

The graph of the function with limit $I = (0, 1)$ is shown below,

Step 3. The values on I and J.

On I, $f$ has a global maximum at $x = 0$ and a global minimum at $x = \frac{5}{\sqrt{e}}$ .

Now,

$\lim_{f (x) \to \infty} = \lim_{x \to \infty} x^{2} \ln (0.2 x) = \infty$

The graph of the function for $J = (0, \infty)$ is shown below,

On $J$ , $f$ has no global minimum and no global maximum.

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Determine the local extrema of a function.
$f (x) = x^{2} \ln (0.2 x), I = (0, 4], J = (0, \infty)$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. The critical point.

Step 3. The values on I and J.

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Determine the local extrema of a function.f(x)=x2ln0.2x,I=(0,4],J=(0,∞)

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Step by step solution

Step 1. Given Information.

Step 2. The critical point.

Step 3. The values on I and J.

One App. One Place for Learning.

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Determine the local extrema of a function.
$f (x) = x^{2} \ln (0.2 x), I = (0, 4], J = (0, \infty)$