Problem 17
Question
(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=x^{1 / 3}-1 $$
Step-by-Step Solution
Verified Answer
a) The function \(f(x) = x^{\frac{1}{3}} - 1\) is increasing in the entire domain \((-\infty, +\infty)\).
b) There are no relative maxima or relative minima for this function.
1Step 1: Find the first derivative of the function
To find the first derivative of the function \(f(x) = x^{\frac{1}{3}} - 1\), apply the power rule of differentiation:
$$
f'(x) = \frac{1}{3}x^{\frac{1}{3} - 1} = \frac{1}{3}x^{-\frac{2}{3}}
$$
2Step 2: Find the critical points
Identify the critical points by finding the values of x for which the first derivative is equal to zero or undefined:
$$
\frac{1}{3}x^{-\frac{2}{3}} = 0
$$
However, there are no values of x for which this equation is true. The first derivative is undefined when \(x = 0\), so the only critical point is \(x = 0\).
3Step 3: Analyze the intervals for increasing and decreasing
Now, we will examine the behavior of the function in the intervals \((-\infty, 0)\) and \((0, +\infty)\):
1. When \(x < 0\), \(f'(x) = \frac{1}{3}x^{-\frac{2}{3}} > 0\), so the function is increasing.
2. When \(x > 0\), \(f'(x) = \frac{1}{3}x^{-\frac{2}{3}} > 0\), so the function is increasing again.
Upon analyzing the intervals, we find that the function is always increasing.
4Step 4: Identify the relative maxima and relative minima
Since the first derivative is always positive (except at \(x = 0\)), and there's no sign change in the intervals around the critical point, there are no relative maxima or relative minima for this function.
So our final results are:
a) The function is increasing in the whole domain, which is \((-\infty, +\infty)\).
b) There are no relative maxima or relative minima for the function.
Key Concepts
First Derivative TestIncreasing and Decreasing IntervalsCritical Points in Calculus
First Derivative Test
Understanding when a function reaches its peaks and valleys is an integral part of calculus, and the First Derivative Test is a powerful tool that aids in this exploration. Essentially, this test uses the first derivative of a function to determine where its graph has relative maxima or minima.
Using the function from our exercise, \( f(x) = x^{1/3} - 1 \), we first find the derivative, which gives us the rate of change of the function. If the derivative changes from positive to negative at a critical point, it suggests a peak, or a relative maximum. If the change is from negative to positive, the function has a trough, or a relative minimum.
However, in our case, the first derivative \( f'(x) = \frac{1}{3} x^{-2/3} \) remains positive throughout the domain, except at \( x = 0 \) where it's undefined. Since there is no change in sign around the critical point, the First Derivative Test tells us that there is no relative maximum or minimum for the function \( f \).
Using the function from our exercise, \( f(x) = x^{1/3} - 1 \), we first find the derivative, which gives us the rate of change of the function. If the derivative changes from positive to negative at a critical point, it suggests a peak, or a relative maximum. If the change is from negative to positive, the function has a trough, or a relative minimum.
However, in our case, the first derivative \( f'(x) = \frac{1}{3} x^{-2/3} \) remains positive throughout the domain, except at \( x = 0 \) where it's undefined. Since there is no change in sign around the critical point, the First Derivative Test tells us that there is no relative maximum or minimum for the function \( f \).
Increasing and Decreasing Intervals
A function's journey can be seen as a hike through hills and valleys, and by identifying its increasing and decreasing intervals, we can trace the path it takes. When a function's derivative is positive over an interval, it means that the function is climbing up or increasing. Conversely, if the derivative is negative, it is on a descent or decreasing.
Following our function \( f(x) = x^{1/3} - 1 \), we can infer from its derivative \( f'(x) \) that it is increasing over two intervals: between negative infinity and zero, and from zero to positive infinity. In both ranges, the derivative remains positive, implying a steady ascent with no dips. This tells us that the function constantly increases without any downward segments.
Following our function \( f(x) = x^{1/3} - 1 \), we can infer from its derivative \( f'(x) \) that it is increasing over two intervals: between negative infinity and zero, and from zero to positive infinity. In both ranges, the derivative remains positive, implying a steady ascent with no dips. This tells us that the function constantly increases without any downward segments.
Critical Points in Calculus
Much like the key moments in a story, in calculus, critical points are where the plot of a function can take a turn. These critical points occur where a function's derivative is either zero or undefined. At these points, we might find a peak, a valley, or a leveling off in the function's graph—locations of potential relative maxima or minima.
Our exercise involves the function \( f(x) = x^{1/3} - 1 \). When hunting for its critical points, we discover that the derivative \( f'(x) \) is never zero but is undefined at \( x = 0 \). This single critical point becomes a subject of interest. However, as our derivative doesn't change signs around \( x = 0 \), this point doesn't signify a relative maximum or minimum but is rather just a point where the function flattens momentarily in its otherwise consistent rise.
Our exercise involves the function \( f(x) = x^{1/3} - 1 \). When hunting for its critical points, we discover that the derivative \( f'(x) \) is never zero but is undefined at \( x = 0 \). This single critical point becomes a subject of interest. However, as our derivative doesn't change signs around \( x = 0 \), this point doesn't signify a relative maximum or minimum but is rather just a point where the function flattens momentarily in its otherwise consistent rise.
Other exercises in this chapter
Problem 17
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