Question

Identify the critical points. Then use (a) the First Derivative Test and (if possible) (b) the Second Derivative Test to decide which of the critical points give a local maximum and which give a local minimum. $$ f(x)=\frac{\ln (x+2)}{x+2} $$

Short answer

The function has a local maximum at \( x = e-2 \).

Step-by-step solution

Find the First Derivative

First, we need to find the derivative of the function \( f(x) = \frac{\ln(x+2)}{x+2} \). We use the quotient rule \( \frac{d}{dx} \left[ \frac{u}{v} \right] = \frac{u'v - uv'}{v^2} \). Let \( u = \ln(x+2) \) and \( v = x+2 \):\( u' = \frac{1}{x+2} \, (u = \ln(x+2)), \quad v' = 1 \, (v = x+2) \)Apply the quotient rule:\( f'(x) = \frac{\frac{1}{x+2} \cdot (x+2) - \ln(x+2) \cdot 1}{(x+2)^2} = \frac{1 - \ln(x+2)}{(x+2)^2}\)

Find the Critical Points

A critical point occurs where the derivative \( f'(x) \) is zero or undefined. Since we found that:\( f'(x) = \frac{1 - \ln(x+2)}{(x+2)^2}\)Set the numerator equal to zero:\( 1 - \ln(x+2) = 0 \)Solving for \(x\):\( \ln(x+2) = 1 \)Exponentiate both sides:\( x+2 = e^1 \)\( x = e - 2 \)The critical points are where \( x = e - 2 \). The denominator \((x+2)^2\) is always positive except where \(x = -2\), hence no critical points from undefined derivative aside \(x > -2\).

Apply the First Derivative Test

To apply the First Derivative Test, examine the sign of \( f'(x) \) to the left and right of the critical point \( x = e - 2 \).For \( x < e - 2 \), choose \( x = 0 \), then \( \ln(x+2) = \ln(2) < 1 \), so \( f'(x) > 0 \).For \( x > e - 2 \), choose \( x = 1 \), then \( \ln(x+2) = \ln(3) > 1 \), so \( f'(x) < 0 \).This change from positive to negative indicates a local maximum at \( x = e - 2 \).

Apply the Second Derivative Test

To further verify, we use the Second Derivative Test. First, find the second derivative \( f''(x) \) by differentiating \( f'(x) = \frac{1 - \ln(x+2)}{(x+2)^2} \).Differentiate using the quotient rule again:Let \( g = 1 - \ln(x+2) \), \( h = (x+2)^2 \)\( g' = -\frac{1}{x+2}, \quad h' = 2(x+2)\)The second derivative is:\( f''(x) = \frac{(-1/(x+2))\cdot (x+2)^2 - (1 - \ln(x+2))\cdot 2(x+2)}{(x+2)^4}\)Simplifying and substituting \( x = e-2 \), \( f''(x) = - \frac{(e-2+2)^2 + 2(e-1)(e-2+2)}{(e)^4} < 0 \)This confirms a local maximum at \( x = e-2 \) as \( f''(x) < 0 \). Therefore, the point \( x = e-2 \) is a local maximum.

Key concepts

First Derivative Test

The First Derivative Test is a handy tool used in calculus to determine whether a critical point is a local maximum or minimum. A critical point is found where the first derivative of a function, denoted as \( f'(x) \), is zero or undefined. Here is how the test works:

• Arrange intervals around the critical point and
• examine the sign of \( f'(x) \) on each interval.

If the derivative changes from positive to negative as it passes through the critical point, then you have a local maximum at that point. Conversely, if it changes from negative to positive, the point is a local minimum. For the function given by \( f(x) = \frac{\ln(x+2)}{x+2} \), the critical point was found to be \( x = e - 2 \). By choosing values less than \( e - 2 \) and greater than \( e - 2 \), you can determine the behavior of \( f'(x) \):

• For \( x < e - 2 \), \( f'(x) > 0 \), indicating increasing behavior.
• For \( x > e - 2 \), \( f'(x) < 0 \), showing decreasing behavior.

Thus, the function increases to the left of \( e - 2 \) and decreases to the right, indicating that \( x = e - 2 \) is a local maximum.

Second Derivative Test

The Second Derivative Test provides another way to determine whether a critical point is a local maximum or minimum. This test examines the second derivative of the function, \( f''(x) \), at the critical point. The steps to apply this test are simple:

• Find \( f''(x) \), the second derivative of the function.
• Evaluate \( f''(x) \) at the critical point.

If \( f''(x) > 0 \) at the critical point, the point is a local minimum as the graph shows concave up behavior. If \( f''(x) < 0 \), then it is a local maximum due to the concave down behavior of the function. In our example of \( f(x) = \frac{\ln(x+2)}{x+2} \), the second derivative evaluated at \( x = e - 2 \) is negative \( f''(e-2) < 0 \).This means that the function graph is curving downwards at this point, confirming the local maximum we found using the First Derivative Test.

Quotient Rule in Calculus

The Quotient Rule is essential in calculus for finding the derivative of a function that is the ratio of two differentiable functions. This rule states:\[\frac{d}{dx} \left[ \frac{u}{v} \right] = \frac{u'v - uv'}{v^2}\]Here, \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their respective derivatives. The rule ensures that you differentiate the numerator and denominator separately before combining them using the formula. In practice, you:

• Calculate the derivative of the numerator \( u' \) while keeping the denominator unchanged.
• Subtract the product of the original numerator \( u \) and the derivative of the denominator \( v' \).
• Divide the entire expression by the square of the denominator \( v^2 \).

Applying the Quotient Rule to the function \( f(x) = \frac{\ln(x+2)}{x+2} \) allowed us to find the first derivative, \( f'(x) = \frac{1 - \ln(x+2)}{(x+2)^2} \), critical for locating critical points and analyzing the function's behavior.