Problem 27 - Essential Calculus

Question

Find the critical numbers of \(f\) (if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results. $$ f(x)=x^{1 / 3}+1 $$

Step-by-Step Solution

Answer

The critical number of the function \(f(x) = x^{1/3} + 1\) is 0. The function is decreasing on the interval (-∞, 0) and increasing on the interval (0, +∞). The function has a relative minimum at x=0.

Step 1: Identification of the function and its derivative

Firstly, identify the given function \(f(x) = x^{1/3} + 1\). Next, find the derivative of the function \(f'(x)\). Using the power rule for differentiation we find that \(f'(x) = \frac{1}{3x^{2/3}}\).

Step 2: Find Critical Numbers

Critical numbers of a function are found by setting the derivative equal to zero or when the derivative is undefined. Notice here as \(x\) approaches 0, the derivative \(f'(x) = \frac{1}{3x^{2/3}}\) will be undefined. Thus, 0 is a critical number.

Step 3: Determine Increasing and Decreasing Intervals

If \(f'(x) > 0\), the function is increasing and if \(f'(x) < 0\), the function is decreasing. Test for \(f'(x)\) in intervals split by the critical number 0 would be (-∞, 0), (0, +∞). You will see that for \(x > 0\) (i.e., in (0, +∞)), \(f'(x) > 0\) thus function is increasing. And for \(x < 0\) (i.e., in (-∞, 0)), \(f'(x) < 0\) thus function is decreasing.

Step 4: Identification of Extrema

Since the function decreases on (-∞, 0) and increases on (0, +∞), there is a relative minimum at x=0. By examining the left and right side of the critical number to determine whether the function is decreasing or increasing we can infer whether a pointer is a relative maximum, minimum or neither.

Step 5: Graphical Representation

Lastly, confirm these findings by graphing the function using a graphing utility. The relative minimum at x=0, along with an increasing function on (0, +∞) and a decreasing function on (-∞, 0) should be observable.

Key Concepts

Derivative of a Function

In calculus, the derivative of a function is an essential concept that represents the rate at which the function's value changes. Whenever we have a function like
\( f(x) = x^{1/3} + 1 \),
the derivative helps us understand how this function behaves as we change
\( x \). To find the derivative, we employ various rules of differentiation such as the power rule. For our function, using the power rule,
\( f'(x) = \frac{d}{dx}(x^{1/3}) + \frac{d}{dx}(1) \)
leads to
\( f'(x) = \frac{1}{3x^{2/3}} \).
Calculating the derivative is the first step in finding critical numbers, determining where the graph increases or decreases, and locating relative extrema. Without derivatives, none of these analyses would be possible.

Increasing and Decreasing Intervals

The intervals where a function is increasing or decreasing tell us a lot about its graph. After finding the derivative, like
\( f'(x) = \frac{1}{3x^{2/3}} \)
for our function, we determine these intervals by looking at the sign of
\( f'(x) \).
If
\( f'(x) > 0 \),
the function is climbing up, which means it's increasing. Conversely, if
\( f'(x) < 0 \),
the function is going down, thus decreasing. In our case,
\( f'(x) > 0 \)
for
\( x > 0 \), meaning the function
\( f \)
is increasing in the interval
\( (0, +\infty) \).
Similarly,
\( f'(x) < 0 \)
for
\( x < 0 \),
indicating decreasing behavior on
\( (-\infty, 0) \). This analysis helps us sketch the function's graph more accurately, and understand where the function is gaining or losing value.

Relative Extrema

Relative extrema refer to the high and low points—peaks and valleys—of a function on certain intervals. They are 'relative' because they are not necessarily the highest or lowest points overall, but they are the most extreme within a nearby range of
\( x \)-values.

Identifying Relative Extrema Using Derivatives

After finding the derivative, like
\( f'(x) = \frac{1}{3x^{2/3}} \)
for our function, critical numbers where
\( f'(x) = 0 \)
or is undefined, create potential locations for these extrema. In our step by step solution, the function has a critical number at
\( x = 0 \),
since the derivative does not exist at that point. By examining the sign of
\( f'(x) \)
around this critical number, we can determine the type of extremum it is. Since the function
\( f \)
is decreasing on
\( (-\infty, 0) \)
and increasing on
\( (0, +\infty) \),
it implies that there is a relative minimum at
\( x = 0 \). Graphing utilities can confirm this finding visually by showing the dip in the function at this point.