Question

Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible ) whether they correspond to local maxima or local minima. $$f(x)=\frac{e^{x}}{x+1}$$

Short answer

Question: Determine if the critical points of the function \(f(x) = \frac{e^x}{x+1}\) are local maxima or local minima using the Second Derivative Test. Answer: The critical point \(x=0\) corresponds to a local minimum.

Step-by-step solution

Find the first derivative of the function

To find the first derivative, we will make use of the quotient rule: $$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}$$ In this case, we have \(u=e^x\) and \(v=x+1\). So the first derivative of our function will be: $$f'(x) = \frac{(x + 1)(\frac{d}{dx}(e^x)) - e^x(\frac{d}{dx}(x+1))}{(x + 1)^2}$$ $$f'(x) = \frac{(x + 1)(e^x) - e^x(1)}{(x + 1)^2}$$ Simplify it: $$f'(x) = \frac{xe^x}{(x + 1)^2}$$

Find the critical points

A critical point occurs when \(f'(x) = 0\) or \(f'(x)\) is undefined. For our function \(f'(x)=\frac{xe^x}{(x+1)^2}\), there is no value of \(x\) for which the first derivative is undefined, so we'll set \(f'(x)=0\) and solve for x: $$\frac{xe^x}{(x + 1)^2} = 0$$ As the exponential function (\(e^x\)) is always positive, we have: $$xe^x = 0$$ We find that \(x=0\) is a critical point.

Find the second derivative of the function

Now we need to find the second derivative of the function, \(f''(x)\), to apply the Second Derivative Test. We will again use the quotient rule on \(f'(x) = \frac{xe^x}{(x+1)^2}\): $$f''(x) = \frac{(x + 1)^2(\frac{d}{dx}(xe^x)) - xe^x(\frac{d}{dx}((x+1)^2))}{(x + 1)^4}$$ We have two terms to differentiate: - The derivative of \(xe^x\) using the product rule: $$(xe^x)' = x'(e^x) + x(e^x)' = e^x + xe^x$$ - The derivative of \((x+1)^2\): $$(x+1)^2'=2(x+1)$$ So, $$f''(x) = \frac{(x + 1)^2(e^x + xe^x) - xe^x(2(x+1))}{(x + 1)^4}$$

Use the Second Derivative Test

We will now apply the Second Derivative Test to the critical point \(x=0\): $$f''(0) = \frac{(0 + 1)^2(e^0 + 0(e^0)) - 0(e^0)(2(0+1))}{(0 + 1)^4}$$ $$f''(0) = \frac{1}{1}$$ Since \(f''(0) > 0\), the Second Derivative Test tells us that the critical point \(x=0\) corresponds to a local minimum.

Key concepts

Second Derivative Test

The Second Derivative Test is a straightforward way to determine the nature of critical points for a function. Once you've found where the first derivative equals zero or is undefined, known as critical points, you can use the second derivative to decide if these points are local minima or maxima.

• If the second derivative at the critical point is positive, the function is concave up, indicating a local minimum.
• If the second derivative at the critical point is negative, the function is concave down, indicating a local maximum.
• If the second derivative equals zero, the test is inconclusive, and further analysis or a different method is needed.

In our exercise, after finding that the critical point is at x=0, we applied the Second Derivative Test and found that the function's second derivative was positive, verifying a local minimum at x=0.

First Derivative

The first derivative of a function measures its instantaneous rate of change. It's calculated through techniques such as the power rule, product rule, or, as seen in our exercise, the quotient rule.
A crucial step in solving calculus problems, the first derivative tells us where the function is increasing or decreasing:

• If the first derivative is positive, the function is increasing.
• If it is negative, the function is decreasing.
• Critical points occur where the first derivative is zero or undefined.

In our function, the first derivative was found using the quotient rule resulting in \( f'(x) = \frac{xe^x}{(x + 1)^2} \). Solving \( f'(x) = 0 \) gave us the critical point at x=0.

Exponentiation

Exponentiation involves raising a number to the power of another and, for functions, often involves variables as exponents. The exponential function \( e^x \) is particularly noteworthy because it is its own derivative. This characteristic makes it vital in calculus.
With \( e^x \), you find:

• The base (e) is a constant approximately equal to 2.71828.
• The exponential function grows faster than polynomial or logarithmic functions.
• As seen in our function, \( e^x \) was a critical component in both the first and second derivatives.

The nature of exponentials makes them useful in modeling various real-world processes involving growth or decay.

Quotient Rule

The Quotient Rule is essential when differentiating functions expressed as fractions. The formula is used when one function is divided by another and is given by:\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2}\]where \( u \) and \( v \) are functions of \( x \). You apply this rule to differentiate complex rational functions accurately.
In our problem, \( f(x) = \frac{e^x}{x+1} \), this rule helped derive the first and second derivatives efficiently.

• Step 1: Differentiate the numerator \( e^x \).
• Step 2: Differentiate the denominator \( x+1 \).
• Step 3: Plug into the Quotient Rule formula.

By following these steps, you correctly assess the function's behavior and find important properties like critical points.