Locate and classify all extrema for the function f(x)= 4\ln x - \frac{1}{2}x^2.

Question:

Locate and classify all extrema for the function {eq}f(x)= 4\ln x - \frac{1}{2}x^2. {/eq}

Extreme Values

To find all extrema of a function {eq}\displaystyle y=f(x) {/eq}, we will look for the absolute extreme values and relative extreme values.

The absolute extreme values are the maximum and minimum values of the function on the entire domain, if they exist.

The relative extreme values of a function are the values of a function calculated on its critical points, such that the derivative function changes signs at those critical points.

The critical points are the points {eq}\displaystyle c, {/eq} from the domain of the function such that

{eq}\displaystyle f'(c)= 0, \text{ or } f'(c) - DNE {/eq}

The absolute extrema will be chosen by evaluating the function in the critical points and the end points of the domain and retain the maximum and minimum value.

Differentiating functions, we may need the following formulae

{eq}\displaystyle \frac{d}{dx}(x^n)=nx^{n-1},\\ \displaystyle \frac{d}{dx}\left[af(x)+bg(x)\right] =a\frac{d}{dx}\left(f(x)\right)+b\frac{d}{dx}\left(g(x)\right), a,b - \text{ constants}. {/eq}

Answer and Explanation:

To find all extrema of the function {eq}\displaystyle f(x)=4\ln x-\frac{1}{2}x^2, {/eq} we need to establish the domain first.

The domain is given by the restriction given by the logarithmic function,

{eq}\displaystyle (0,\infty). {/eq}

Because there are no end points on the interval, the absolute extreme will be among the local extreme.

To find the local extreme values, we need the derivative function.

{eq}\displaystyle f'(x)=\frac{d}{dx}\left( 4\ln x-\frac{1}{2}x^2\right) =\frac{4}{x}-x=\frac{4-x^2}{x}. {/eq}

The critical points are obtain from the following conditions

{eq}\displaystyle \begin{align*}f'(x) =0 \text{ or } f'(x) - DNE \\ \frac{4-x^2}{x}=0 \text{ or } \frac{4-x^2}{x} - DNE \\ 4-x^2=0 \text{ or }x=0 \\ x=-2 \notin(0,\infty), 2 \in(0,\infty) \text{ or }x=0 \notin(0,\infty) \\ \end{align*} {/eq}

So, the only critical point is {eq}\displaystyle x=2. {/eq}

To find if it is maximum or minimum, we look at the sign of the derivative

{eq}\displaystyle \begin{array}{ccc} \text{ critical points } & & 2 & \\ \hline x & 1 & 2& 3\\ \hline \text{ signs of } f'(x) & +& 0&- \end{array} {/eq}

Therefore, the critical point {eq}\displaystyle x=2 \text{ is a local and absolute maximum, and} \boxed{\text{ the maximum value is } f(2)= 4\ln 2 - 2}. {/eq}


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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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