Problem 24
Question
In Exercises 21 through 24 , find all critical numbers for the given function \(f(x)\) and use the second derivative test to determine which (if any) critical points are relative maxima or relative minima. $$ f(x)=\frac{1}{x}-\frac{1}{x+3} $$
Step-by-Step Solution
Verified Answer
The critical point is at \( x = -\frac{3}{2} \) and it's a relative maximum.
1Step 1 - Find the first derivative
To find the critical numbers, we first need to find the first derivative of the function. Given \[ f(x) = \frac{1}{x} - \frac{1}{x+3} \], use the quotient rule twice.\[ f'(x) = \frac{d}{dx}\left(\frac{1}{x}\right) - \frac{d}{dx}\left(\frac{1}{x+3}\right) \]. Apply the quotient rule: \[ \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \] and \[ \frac{d}{dx}\left(\frac{1}{x+3}\right) = -\frac{1}{(x+3)^2} \]. Combining these, we get:\[ f'(x) = -\frac{1}{x^2} + \frac{1}{(x+3)^2} \].
2Step 2 - Set the first derivative to 0
Set the first derivative equal to zero to find the critical points:\[ -\frac{1}{x^2} + \frac{1}{(x+3)^2} = 0 \]. This can be rewritten as:\[ \frac{1}{(x+3)^2} = \frac{1}{x^2} \]. Cross-multiplying, we get:\[ x^2 = (x+3)^2 \].
3Step 3 - Solve for x
Expand and simplify the equation: \[ x^2 = x^2 + 6x + 9 \]. Cancel out \(x^2\) from both sides:\[ 0 = 6x + 9 \]. Solving for \(x\), we get:\[ x = -\frac{3}{2} \]. This is a critical point of the function.
4Step 4 - Find the second derivative
To use the second derivative test, we need the second derivative of the function. The first derivative was \[ f'(x) = -\frac{1}{x^2} + \frac{1}{(x+3)^2} \]. Differentiate again:\[ f''(x) = \frac{2}{x^3} - \frac{2}{(x+3)^3} \].
5Step 5 - Evaluate the second derivative at the critical point
Substitute \( x = -\frac{3}{2} \) into the second derivative to determine concavity:\[ f''\left(-\frac{3}{2}\right) = \frac{2}{\left(-\frac{3}{2}\right)^3} - \frac{2}{\left(-\frac{3}{2} + 3\right)^3} \]. Simplify each term separately:\[ f''\left(-\frac{3}{2}\right) = \frac{2}{-\frac{27}{8}} - \frac{2}{\frac{27}{8}} = -\frac{16}{27} - \frac{16}{27} = -\frac{32}{27} \].
6Step 6 - Interpret the second derivative test result
Since \( f''\left(-\frac{3}{2}\right) < 0 \), the critical point \( x = -\frac{3}{2} \) is a point of relative maximum.
Key Concepts
First derivativeSecond derivative testRelative maxima and minimaQuotient rule
First derivative
The first derivative is a fundamental concept in calculus, indicating the rate of change of a function. To find the critical points of a function, we first need to calculate its first derivative. This involves differentiating the function with respect to its variable. In the exercise, we started with the function \( f(x) = \frac{1}{x} - \frac{1}{x+3} \).
Using the quotient rule twice, we derived \( f'(x) = -\frac{1}{x^2} + \frac{1}{(x+3)^2} \). This is known as the first derivative.
Setting the first derivative equal to zero helps us find the critical points. This step is crucial because it tells us where the function's rate of change is zero. For our given function, setting \( -\frac{1}{x^2} + \frac{1}{(x+3)^2} = 0 \) led to the equation \( x^2 = (x+3)^2 \). Solving it, we obtained the critical point \( x = -\frac{3}{2} \).
Using the quotient rule twice, we derived \( f'(x) = -\frac{1}{x^2} + \frac{1}{(x+3)^2} \). This is known as the first derivative.
Setting the first derivative equal to zero helps us find the critical points. This step is crucial because it tells us where the function's rate of change is zero. For our given function, setting \( -\frac{1}{x^2} + \frac{1}{(x+3)^2} = 0 \) led to the equation \( x^2 = (x+3)^2 \). Solving it, we obtained the critical point \( x = -\frac{3}{2} \).
Second derivative test
The second derivative test helps determine the nature of critical points, whether they are relative maxima, minima, or neither. After finding the first derivative, we differentiate it again to obtain the second derivative.
For the function \( f(x) = \frac{1}{x} - \frac{1}{x+3} \), we calculated the first derivative as \( f'(x) = -\frac{1}{x^2} + \frac{1}{(x+3)^2} \).
Next, differentiating again, we found the second derivative to be \( f''(x) = \frac{2}{x^3} - \frac{2}{(x+3)^3} \). By evaluating the second derivative at the critical point \( x = -\frac{3}{2} \), we got \( f''\left(-\frac{3}{2}\right) = -\frac{32}{27} \).
Since \( f''\left(-\frac{3}{2}\right) < 0 \), we concluded that the critical point \( x = -\frac{3}{2} \) is a point of relative maximum. When the second derivative at a critical point is less than zero, it indicates that the function is concave down at that point, confirming the presence of a relative maximum.
For the function \( f(x) = \frac{1}{x} - \frac{1}{x+3} \), we calculated the first derivative as \( f'(x) = -\frac{1}{x^2} + \frac{1}{(x+3)^2} \).
Next, differentiating again, we found the second derivative to be \( f''(x) = \frac{2}{x^3} - \frac{2}{(x+3)^3} \). By evaluating the second derivative at the critical point \( x = -\frac{3}{2} \), we got \( f''\left(-\frac{3}{2}\right) = -\frac{32}{27} \).
Since \( f''\left(-\frac{3}{2}\right) < 0 \), we concluded that the critical point \( x = -\frac{3}{2} \) is a point of relative maximum. When the second derivative at a critical point is less than zero, it indicates that the function is concave down at that point, confirming the presence of a relative maximum.
Relative maxima and minima
Relative maxima and minima are points where a function reaches local highs or lows. A relative maximum occurs where the function reaches a peak, and a relative minimum occurs where it reaches a trough within a certain interval.
After identifying the critical points using the first derivative, we use the second derivative to determine their nature. In our exercise, we found a single critical point at \( x = -\frac{3}{2} \).
By evaluating the second derivative at this point, we saw that \( f''\left(-\frac{3}{2}\right) < 0 \). This negative value indicates that the function is concave down at this critical point, confirming it as a relative maximum. Therefore, the function has a local high at \( x = -\frac{3}{2} \).
After identifying the critical points using the first derivative, we use the second derivative to determine their nature. In our exercise, we found a single critical point at \( x = -\frac{3}{2} \).
By evaluating the second derivative at this point, we saw that \( f''\left(-\frac{3}{2}\right) < 0 \). This negative value indicates that the function is concave down at this critical point, confirming it as a relative maximum. Therefore, the function has a local high at \( x = -\frac{3}{2} \).
Quotient rule
The quotient rule is a method used in calculus to differentiate functions that are divided by each other. It states that if you have a function \( f(x) = \frac{u(x)}{v(x)} \), then its derivative is given by:
\[ f'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{v(x)^2} \]
In our exercise, we used the quotient rule to differentiate each part of the function \( f(x) = \frac{1}{x} - \frac{1}{x+3} \).
For \( \frac{1}{x} \), we applied the quotient rule to obtain \( -\frac{1}{x^2} \). Likewise, for \( \frac{1}{x+3} \), we derived \( -\frac{1}{(x+3)^2} \). Combining these results, we got the first derivative \( f'(x) = -\frac{1}{x^2} + \frac{1}{(x+3)^2} \).
The quotient rule is indispensable when dealing with complex fractions as it provides a systematic way to find the derivative of divided functions.
\[ f'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{v(x)^2} \]
In our exercise, we used the quotient rule to differentiate each part of the function \( f(x) = \frac{1}{x} - \frac{1}{x+3} \).
For \( \frac{1}{x} \), we applied the quotient rule to obtain \( -\frac{1}{x^2} \). Likewise, for \( \frac{1}{x+3} \), we derived \( -\frac{1}{(x+3)^2} \). Combining these results, we got the first derivative \( f'(x) = -\frac{1}{x^2} + \frac{1}{(x+3)^2} \).
The quotient rule is indispensable when dealing with complex fractions as it provides a systematic way to find the derivative of divided functions.
Other exercises in this chapter
Problem 22
In Exercises 21 through 24 , find all critical numbers for the given function \(f(x)\) and use the second derivative test to determine which (if any) critical p
View solution Problem 23
In Exercises 21 through 24 , find all critical numbers for the given function \(f(x)\) and use the second derivative test to determine which (if any) critical p
View solution Problem 25
In Exercises 25 through 28, find the absolute maximum and the absolute minimum values (if any) of the given function on the specified interval. $$ f(x)=-2 x^{3}
View solution Problem 26
In Exercises 25 through 28, find the absolute maximum and the absolute minimum values (if any) of the given function on the specified interval. $$ f(t)=-3 t^{4}
View solution