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Magazine Chapter 4: Problem 29
Find the intervals on which \(f\) is increasing and decreasing. $$f(x)=x^{2 / 3}\left(x^{2}-4\right)$$
Short Answer
Expert verified
The function f(x) is increasing on the interval (0, 鈭�) and decreasing on the interval (-鈭�, 0).
Step by step solution
01
Find the first derivative
First, we need to find the derivative of the function, \(f'(x)\). So, we need the derivative of
$$
f(x)=x^{2 / 3}\left(x^{2}-4\right).
$$
Using product rule, if \(u(x)=x^{2/3}\) and \(v(x)=x^2-4\), we can differentiate both \(u\) and \(v\).
$$
u'(x)=\frac{2}{3}x^{2/3-1},
$$
$$
v'(x)=2x.
$$
Now using product rule to find \(f'(x)\)
$$
f'(x)=u'(x)v(x) + u(x)v'(x)
$$
$$
f'(x) = \frac{2}{3}x^{2/3-1}(x^{2}-4) + x^{2/3}(2x)
$$
$$
f'(x)=\frac{2}{3} x^{-1/3}\left(x^{2}-4\right) + 2x^{5/3}
$$
02
Find critical points
To find the critical points, we must determine where the derivative is either equal to \(0\) or undefined.
$$
f'(x)=\frac{2}{3} x^{-1/3}\left(x^{2}-4\right) + 2x^{5/3}
$$
First, we find the points when \(f'(x) = 0\):
$$
\frac{2}{3} x^{-1/3}\left(x^{2}-4\right) + 2x^{5/3} =0
$$
Second, we find the points when \(f'(x)\) is undefined (when the term \(x^{-1/3}\) is undefined):
$$
x^{-1/3}=x^\frac{-1}{3} \Rightarrow x^{1/3}=0.
$$
Since there is no real value for \(x\) that satisfies this equation, it means \(f'(x)\) only has undefined values when \(x=0\). So, \(x=0\) is a critical point.
03
Test intervals around critical points
We next analyze the sign of \(f'(x)\) in intervals around the critical point \(x=0\) to determine increasing or decreasing behavior.
The intervals are \((-\infty,0)\) and \((0, \infty)\).
To test the interval \((-\infty,0)\), we pick an arbitrary number, say \(-1\):
$$
f'(-1) = \frac{2}{3}(-1)^{-1/3}\left(1-4\right) + 2(-1)^{5/3} <0
$$
So, \(f(x)\) is decreasing in the interval \((-\infty,0)\).
To test the interval \((0, \infty)\), we pick an arbitrary number, say \(1\):
$$
f'(1) = \frac{2}{3}(1)^{-1/3}\left(1^2-4\right) + 2(1)^{5/3} >0
$$
So, \(f(x)\) is increasing in the interval \((0, \infty)\).
04
Conclusion
Based on the signs of \(f'(x)\) in the tested intervals, we can conclude the following behavior for the function:
1. \(f(x)\) is decreasing on the interval \((-\infty,0)\).
2. \(f(x)\) is increasing on the interval \((0,\infty)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
A derivative represents how a function changes as its input changes. It's like measuring the speed of a moving car at each point instead of just knowing the car's general direction.
The derivative of a function gives the slope of that function at any point. Mathematically, if you have a function \(f(x)\), its derivative is often written as \(f'(x)\).
To find the derivative of a product of two functions, we use the product rule. This rule involves differentiating each function separately and then combining the results. For our function \(f(x)=x^{2 / 3}(x^2-4)\), we first differentiate \(u(x)=x^{2/3}\) to get \(u'(x)=\frac{2}{3}x^{-1/3}\), and \(v(x)=x^2-4\) to get \(v'(x)=2x\).
Hence, the derivative \(f'(x)\) is calculated using:
The derivative of a function gives the slope of that function at any point. Mathematically, if you have a function \(f(x)\), its derivative is often written as \(f'(x)\).
To find the derivative of a product of two functions, we use the product rule. This rule involves differentiating each function separately and then combining the results. For our function \(f(x)=x^{2 / 3}(x^2-4)\), we first differentiate \(u(x)=x^{2/3}\) to get \(u'(x)=\frac{2}{3}x^{-1/3}\), and \(v(x)=x^2-4\) to get \(v'(x)=2x\).
Hence, the derivative \(f'(x)\) is calculated using:
- \(f'(x) = u'(x)v(x) + u(x)v'(x)\)
- This method helps find where the function is increasing or decreasing.
Critical Points
Critical points are where a function changes its direction from increasing to decreasing or vice versa. They're like turning points on a road - spots where the journey is about to shift.
Critical points occur where the derivative \(f'(x)\) is zero or undefined. For our example \(f'(x)=\frac{2}{3} x^{-1/3}(x^{2}-4) + 2x^{5/3}\), we check these conditions.
When \(f'(x) = 0\), solve the equation for \(x\). Additionally, check where the derivative can't be evaluated. Here, \(x=0\) makes \(x^{-1/3}\) undefined, pointing to \(x=0\) as a critical point. Critical points highlight important features in the graph of a function.
To summarize:
Critical points occur where the derivative \(f'(x)\) is zero or undefined. For our example \(f'(x)=\frac{2}{3} x^{-1/3}(x^{2}-4) + 2x^{5/3}\), we check these conditions.
When \(f'(x) = 0\), solve the equation for \(x\). Additionally, check where the derivative can't be evaluated. Here, \(x=0\) makes \(x^{-1/3}\) undefined, pointing to \(x=0\) as a critical point. Critical points highlight important features in the graph of a function.
To summarize:
- Identify critical points by setting \(f'(x)=0\) and checking for undefined values.
- Determine the nature (maxima, minima) at these points by analyzing the intervals around them.
Increasing and Decreasing Functions
To determine where a function is increasing or decreasing, analyze the sign of the derivative \(f'(x)\).
If \(f'(x) > 0\), the function is increasing, indicating growth. If \(f'(x) < 0\), it's decreasing, showing decline.
In our case, we checked intervals around the critical point \(x=0\). In the interval \((-\infty,0)\), we picked \(x=-1\) and found \(f'(-1)<0\), which shows that \(f(x)\) is decreasing.
In the interval \((0,\infty)\), we picked \(x=1\) and found \(f'(1)>0\), so \(f(x)\) is increasing there.
Essentially, by testing values in intervals separated by critical points, we determine the behavior:
If \(f'(x) > 0\), the function is increasing, indicating growth. If \(f'(x) < 0\), it's decreasing, showing decline.
In our case, we checked intervals around the critical point \(x=0\). In the interval \((-\infty,0)\), we picked \(x=-1\) and found \(f'(-1)<0\), which shows that \(f(x)\) is decreasing.
In the interval \((0,\infty)\), we picked \(x=1\) and found \(f'(1)>0\), so \(f(x)\) is increasing there.
Essentially, by testing values in intervals separated by critical points, we determine the behavior:
- Decreasing function in \((-\infty,0)\)
- Increasing function in \((0,\infty)\)
