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Magazine Chapter 10: Problem 47
Determine where the function is concave upward and where it is concave downward. $$ f(x)=x^{2}+\ln x^{2} $$
Short Answer
Expert verified
The function \(f(x) = x^2 + \ln(x^2)\) is concave upward on the interval \((-1, 1)\) and concave downward on the intervals \((-\infty, -1)\) and \((1, \infty)\).
Step by step solution
01
Find the first derivative of the function
To find the first derivative of the function, we will use the sum and chain rule.
For the given function, \(f(x) = x^2 + \ln(x^2)\), we'll first find the derivative of \(x^2\) with respect to x and the derivative of \(\ln(x^2)\) with respect to x.
Derivative of \(x^2\) with respect to x:
\(f_1'(x) = 2x\)
Derivative of \(\ln(x^2)\) with respect to x:
\(f_2'(x) = \frac{1}{x^2} \cdot 2x = \frac{2}{x}\)
Using the sum rule, the first derivative of the function is:
\(f'(x) = f_1'(x) + f_2'(x) = 2x + \frac{2}{x}\)
02
Find the second derivative of the function
Now, we will find the second derivative of the function by differentiating the first derivative that we found in step 1.
For the given first derivative, \(f'(x) = 2x + \frac{2}{x}\), we'll find the derivative of \(2x\) with respect to x and the derivative of \(\frac{2}{x}\) with respect to x.
Derivative of \(2x\) with respect to x:
\(f_1''(x) = 2\)
Derivative of \(\frac{2}{x}\) with respect to x:
\(f_2''(x) = - \frac{2}{x^2}\)
Using the sum rule, the second derivative of the function is:
\(f''(x) = f_1''(x) + f_2''(x) = 2 - \frac{2}{x^2}\)
03
Determine the intervals of concave upward and concave downward
Now, we will determine the intervals of concave upward and concave downward by analyzing the sign of the second derivative.
Recall that if \(f''(x) > 0\), the function is concave upward, and if \(f''(x) < 0\), the function is concave downward.
Given the second derivative, \(f''(x) = 2 - \frac{2}{x^2}\), let's find the critical points:
Set \(f''(x) = 0\):
\(2 - \frac{2}{x^2} = 0\)
Solve for x:
\(\frac{2}{x^2} = 2\)
\(x^2 = 1\)
\(x = \pm1\)
Now, we will analyze the sign of the second derivative in each interval:
1. \(x < -1\): In this interval, \(f''(x) < 0\), so the function is concave downward.
2. \(-1 < x < 1\): In this interval, \(f''(x) > 0\), so the function is concave upward.
3. \(x > 1\): In this interval, \(f''(x) < 0\), so the function is concave downward.
04
Write the conclusion
From our analysis, we have found that the given function \(f(x) = x^2 + \ln(x^2)\) is concave upward on the interval \((-1, 1)\) and concave downward on the intervals \((-\infty, -1)\) and \((1, \infty)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Concavity
In calculus, concavity tells us about the shape of the graph of a function. It helps us understand how the graph bends and the direction of its curvature. Concavity is determined by the second derivative of a function.
- If the second derivative, denoted as \(f''(x)\), is positive, the graph of the function is said to be concave upward.
- Conversely, if \(f''(x)\) is negative, the graph is concave downward.
Derivatives
A derivative measures how a function changes as its input changes. It tells us the rate of change or slope of the function at any point. There are different orders of derivatives.
- The first derivative, \(f'(x)\), provides the slope of the tangent line to the function at any given point. It indicates the direction in which the function is moving.
- The second derivative, \(f''(x)\), takes this a step further and shows how the rate of change itself is changing. This is key to understanding concavity.
Concave Upward
When a function's graph is concave upward, it bends like a "U". This means that the slope of the tangent line is increasing. In technical terms, the second derivative \(f''(x)\) is greater than zeroe.g., \(f''(x) > 0\).
This denotes that the function is accelerating upwards, and any tangent lines drawn will lie below the curve, indicating that the curve is opening upwards.
In the given problem, the function \(f(x) = x^2 + \ln(x^2)\) is found to be concave upward in the interval \((-1, 1)\). This means within this range, the graph bends upwards and the second derivative is positive. It's essential to identify such intervals to understand the function's graph behavior accurately.
This denotes that the function is accelerating upwards, and any tangent lines drawn will lie below the curve, indicating that the curve is opening upwards.
In the given problem, the function \(f(x) = x^2 + \ln(x^2)\) is found to be concave upward in the interval \((-1, 1)\). This means within this range, the graph bends upwards and the second derivative is positive. It's essential to identify such intervals to understand the function's graph behavior accurately.
Concave Downward
Contrarily, when a function's graph is concave downward, it resembles an upside-down "U". In this case, the graph curves downwards. Here, the second derivative \(f''(x)\) is less than zero, \(f''(x) < 0\).
This indicates that the rate of change of the slope is decreasing, and tangent lines drawn at any point will tend to lie above the curve, showing that the graph is opening downwards.
In our exercise, the function \(f(x) = x^2 + \ln(x^2)\) is concave downward on the intervals \((-\infty, -1)\) and \((1, \infty)\). Recognizing these intervals allows us to predict behaviors like the graph turning downwards in these ranges. Understanding when a function is concave downward helps in identifying the global and local maxima.
This indicates that the rate of change of the slope is decreasing, and tangent lines drawn at any point will tend to lie above the curve, showing that the graph is opening downwards.
In our exercise, the function \(f(x) = x^2 + \ln(x^2)\) is concave downward on the intervals \((-\infty, -1)\) and \((1, \infty)\). Recognizing these intervals allows us to predict behaviors like the graph turning downwards in these ranges. Understanding when a function is concave downward helps in identifying the global and local maxima.
