Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find, if possible, the (global) maximum and minimum values of the given function on the indicated interval. $$ g(x)=\frac{x^{2}}{x^{3}+32} \text { on }[0, \infty) $$

Step-by-Step Solution

Answer

Global minimum is 0, global maximum is \( \frac{1}{6} \) at \( x = 4 \).

Step 1: Identify the Function and its Interval

We are given the function \( g(x) = \frac{x^2}{x^3 + 32} \) and we need to find the global maximum and minimum on the interval \([0, \infty)\). The function is continuous on this interval and differentiable for \( x \geq 0 \) because the denominator only becomes zero if \( x = -32^{1/3} \), which is negative and outside our interval.

Step 2: Calculate the Derivative

To find critical points where potential extrema occur, we first calculate the derivative of \( g(x) \). Using the quotient rule, \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), we have:\[ u = x^2, \quad v = x^3 + 32 \]\[ u' = 2x, \quad v' = 3x^2 \]Therefore, the derivative is:\[ g'(x) = \frac{(2x)(x^3 + 32) - x^2(3x^2)}{(x^3 + 32)^2} = \frac{2x^4 + 64x - 3x^4}{(x^3 + 32)^2} = \frac{-x^4 + 64x}{(x^3 + 32)^2} \]

Step 3: Find Critical Points

Critical points occur where \( g'(x) = 0 \) or \( g'(x) \) is undefined. Since the denominator \((x^3 + 32)^2\) never becomes zero for \( x \geq 0 \), we solve the numerator for zero:\[ -x^4 + 64x = 0 \]\[ x(-x^3 + 64) = 0 \]This implies \( x = 0 \) or \( x^3 = 64 \). This gives critical points \( x = 0 \) and \( x = 4 \).

Step 4: Evaluate the Function at Critical Points and Endpoints

We evaluate \( g(x) \) at the critical points and the endpoint:\( g(0) = \frac{0^2}{0^3 + 32} = 0 \)\( g(4) = \frac{4^2}{4^3 + 32} = \frac{16}{64+32} = \frac{16}{96} = \frac{1}{6} \)Since \( g(x) \rightarrow 0 \) as \( x \rightarrow \infty \), we also evaluate the limit:\[ \lim_{x \to \infty} g(x) = \lim_{x \to \infty} \frac{x^2}{x^3 + 32} = \lim_{x \to \infty} \frac{1}{x} = 0 \]

Step 5: Conclusion

The values we have evaluated are: \( g(0) = 0 \), \( g(4) = \frac{1}{6} \), and \( \lim_{x \to \infty} g(x) = 0 \). Given this, the global minimum is \( 0 \) occurring at \( x = 0 \) (and as \( x \to \infty \)), and the global maximum is \( \frac{1}{6} \) occurring at \( x = 4 \).

Key Concepts

Critical Points

Critical points are essential in calculus as they help locate potential peaks or troughs of a function. They are values of the variable at which the derivative of a function is zero or undefined. To find these points, one needs to solve the equation where the derivative equals zero. In our function, the derivative was calculated and set to zero: \[ -x^4 + 64x = 0 \]This equation implies:

• If the derivative is zero, then either \( x = 0 \) or \( x^3 = 64 \), giving us \( x = 4 \).

Consequently, critical points are at \( x = 0 \) and \( x = 4 \). These points are candidates for points where the function could achieve its highest or lowest value on the given interval.

Global Maximum and Minimum

Determining the global maximum and minimum values of a function involves evaluating the function at its critical points and endpoints. It shows where the highest and lowest values of the function occur on the defined interval.In our example:

• The function was evaluated at critical points \( x = 0 \) and \( x = 4 \).
• Endpoint behavior was assessed since the interval is \([0, \infty)\).

The evaluations were:

• At \( x = 0 \), the function \( g(0) = 0 \).
• At \( x = 4 \), \( g(4) = \frac{1}{6} \).
• As \( x \to \infty \), \( g(x) \to 0 \).

Thus, the global minimum value is \( 0 \) (at \( x = 0 \) and as \( x \to \infty \)) and the global maximum is \( \frac{1}{6} \) occurring at \( x = 4 \).These are crucial in understanding the behavior of functions over specific intervals.

Quotient Rule

The Quotient Rule is a method in calculus for finding the derivative of a ratio of two differentiable functions. For a function \( g(x) = \frac{u(x)}{v(x)} \), where both \( u \) and \( v \) are differentiable, the quotient rule states:\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\]Applying this rule to our function, \( u = x^2 \) and \( v = x^3 + 32 \), resulted in:

• \( u' = 2x \)
• \( v' = 3x^2 \)
• Derivative: \( \frac{(2x)(x^3 + 32) - x^2(3x^2)}{(x^3 + 32)^2} \)

Simplifying gives:\[g'(x) = \frac{-x^4 + 64x}{(x^3 + 32)^2}\]Using the quotient rule helps efficiently tackle derivatives of fraction-like functions, including multi-step calculations.

Derivative Calculation

Derivative calculation allows us to understand a function's rate of change and identify critical points of interest. For the function \( g(x) = \frac{x^2}{x^3 + 32} \), finding its derivative was essential to locate potential maxima and minima.

• The derivative was calculated using the Quotient Rule, showing where the function's slope is either zero or does not exist.
• Once derived, it was set to zero to solve for critical points: \(-x^4 + 64x = 0\).

By deciphering the derivative, one gains insight into the function's slope across different x-values, revealing points of potential global maxima or minima.Understanding how to derive and interpret functions is fundamental in calculus, providing deeper comprehension and analytical ability to tackle more complex scenarios.