Question

In Exercises, find the absolute extrema of the function on the interval $[0,\mathrm{\infty })$. $$f(x)=8-\frac{4x}{x^2+1}$$

Short answer

The absolute extrema of the function on the interval [0, infinity) is at x=0 which is 8

Step-by-step solution

Compute the derivative of the function

First step involves finding the derivative of the function using the quotient rule : $$f^{\prime }(x)=\frac{(x^2+1)\cdot 4-4x\cdot (2x)}{{(x^2+1)}^2}$$ After simplifying this equation we get $f^{\prime }(x)=\frac{4}{(x^2+1{)}^2}$

Find the critical points of the function

The critical points of the function are found by setting the derivative equal to zero and solving for x. However, as the derivative is always positive for real values of x and never equals 0, there aren't any critical values to be found within the given domain.

Evaluate the function at the end-points

Given that there are no critical points, we must evaluate the function at the only end-point of the interval actually reachable, which is x=0, as $x=\mathrm{\infty }$ is an unreachable point. Substituting x = 0 into the function yields: $f(0)=8-\frac{4\cdot 0}{0^2+1}=8$

Determine absolute extrema

As we only have one tangible point to evaluate (x=0), the function has its minimum and maximum at this point which is 8 as x gets bigger, the function approaches 8, but will never be bigger. Due to this we have the absolute minimum and maximum at x=0, which has the value of 8.

Key concepts

Absolute Extrema

Absolute extrema refer to the highest or lowest values a function takes on an interval. In simpler terms, it's the maximum and minimum values of the function within a certain range.
To determine absolute extrema, you typically check:

• Critical points, where the derivative is zero or undefined
• Endpoints of the interval

The highest value is the absolute maximum, and the lowest is the absolute minimum.
In this exercise, since the function's derivative does not vanish, the extrema are evaluated at the endpoint of the interval. At x=0, the function has a value of 8, indicating it acts as both the absolute minimum and maximum.

Derivative

Derivatives are a fundamental concept in calculus. They represent the rate of change of a function with respect to a variable. When you take the derivative of a function, you are essentially finding its slope at any given point.
For the given function:

• The derivative is found using the quotient rule.
• The quotient rule is useful when you have a ratio of two functions: $\text{Math input error}$.
• The formula is: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\cdot u^{\prime }-u\cdot v^{\prime }}{v^2}$.

After applying this rule and simplifying, the derivative $f^{\prime }(x)=\frac{4}{(x^2+1{)}^2}$ is obtained.
This derivative doesn't reach zero within the interval meaning no critical points are formed.

Critical Points

Critical points are where the derivative of a function is either zero or undefined. These points are key to determining where extrema can occur.
To find critical points, set the derivative equal to zero and solve for the variable. In this problem, however, $f^{\prime }(x)=\frac{4}{(x^2+1{)}^2}$ is always positive for real numbers, meaning it never equals zero.
Additionally, as the function's derivative does not become undefined within the given interval, this function has no critical points here. Without these critical points, the potential extrema are restricted to the behavior of the function at the endpoints of the interval.

Interval Notation

Interval notation is a way of writing subsets of real numbers that describe where a function is evaluated. It's efficient for denoting the domain or range.
For example, $[0,\mathrm{\infty })$ captures all numbers from 0 to infinity. The bracket $[$ means that the endpoint 0 is included, while the parenthesis $)$ indicates infinity is not a reachable endpoint.
When solving for extrema, you must always consider how the function behaves on the specified interval. In this problem, since we are analyzing $f(x)$ on $[0,\mathrm{\infty })$, we inspect the function starting from 0 and understand that as x increases to infinity, the output behavior of the function should also be considered. Using interval notation allows for a clear and concise representation of where solutions are applicable.