Problem 20 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find all relative extrema of the function. Use the Second- Derivative Test when applicable. $$ f(x)=\frac{18}{x^{2}+3} $$

Step-by-Step Solution

Answer

The function \(f(x)=\frac{18}{x^{2}+3}\) has a relative maximum at \(x = 0\).

Step 1: Find the Derivative of the Function

The derivative \(f'(x)\) of the function using the quotient rule is given as \[f'(x) = \frac{(x^2+3)(0) -18(2x)}{(x^2+3)^2}\] Simplify the derivative to get: \[f'(x) = \frac{-36x}{(x^2 + 3)^2}\]

Step 2: Find the Critical Points

The critical points of the function are obtained when the derivative of the function is equal to zero or undefined. Henceforth, set \(\frac{-36x}{(x^2 + 3)^2}=0\). From this equation, \(x = 0\) is a critical point of this function.

Step 3: Use Second-Derivative Test

Find the second derivative \(f''(x)\) of the function by differentiating \(f'(x)\). This turns out to be \[f''(x)= \frac{-36(x^4+10x^2+9)}{(x^2+3)^4}\] Evaluating \(f''(x)\) at \(x = 0\) gives \(f''(0) = -36\). Since \(f''(0)<0\), by the Second-Derivative Test, \(f(x)\) has a local maximum at \(x = 0\).

Key Concepts

Critical Points

When exploring a function's behavior, finding the critical points is a crucial first step to identify potential relative extrema. Critical points occur where the derivative of a function either equals zero or is undefined. These points signify where the function might have a local minimum, maximum, or a point of inflection.

In our exercise, the function is given as \(f(x)=\frac{18}{x^{2}+3}\). To find the critical points, we first identified the derivative, \(f'(x)\), and then solved \(f'(x)=0\). This led us to the expression \(\frac{-36x}{(x^2 + 3)^2}=0\). Since the denominator is always positive and non-zero for real \(x\), the equation simplifies to \(-36x = 0\).

Thus, the only critical point is at \(x=0\). This doesn't automatically tell us whether it's a relative maximum or minimum, but it indicates where to test further with additional methods like the Second-Derivative Test.

Derivative

The derivative of a function is a powerful tool that measures how the function value changes as its input changes. Think of it as giving us the slope of the tangent line to the function’s curve at any point. For our function \(f(x)=\frac{18}{x^{2}+3}\), we use the quotient rule to find \(f'(x)\).

The quotient rule is applied to functions in the form of \(\frac{u}{v}\), with the derivative given by \(\frac{u'v - uv'}{v^2}\). For \(f(x)\), \(u = 18\) (a constant) and \(v = x^2 + 3\). We calculate \(u' = 0\) and \(v' = 2x\), leading to:
\[f'(x) = \frac{(x^2+3)(0) -18(2x)}{(x^2+3)^2} = \frac{-36x}{(x^2 + 3)^2}\]

This derivative assists us in finding the critical points and understanding changes in the function's behavior. It also confirms that the derivative equals zero when \(x = 0\).

Second-Derivative Test

Once critical points are identified, the Second-Derivative Test helps determine if these points are local maxima, minima, or neither. The procedure involves calculating the second derivative of the function and checking its value at each critical point.

In our function example, we calculated the second derivative, \(f''(x)\), from \(f'(x) = \frac{-36x}{(x^2 + 3)^2}\). Differentiating this gives:
\[f''(x) = \frac{-36(x^4+10x^2+9)}{(x^2+3)^4}\]

Evaluating this at our critical point \(x = 0\), we find \(f''(0) = -36\).

The rule for the Second-Derivative Test is straightforward:

• If \(f''(x) > 0\), the function has a local minimum at \(x\).
• If \(f''(x) < 0\), the function has a local maximum at \(x\).
• If \(f''(x) = 0\), the test is inconclusive.

Since \(f''(0) = -36 < 0\), the function \(f(x)\) has a local maximum at \(x = 0\). This confirms the critical point is indeed a relative maximum.