Problem 26 - Calculus Early Transcendentals

Question

Use the given derivative to find all critical points of \(f\) and at each critical point determine whether a relative maximum, relative minimum, or neither occurs. Assume in each case that \(f\) is continuous everywhere. $$ f^{\prime}(x)=4 x^{3}-9 x $$

Step-by-Step Solution

Answer

Critical points: \(x = 0, x = \frac{3}{2}, x = -\frac{3}{2}\); max at \(x = 0\), min at \(x = \pm\frac{3}{2}\).

Step 1: Find Critical Points

A critical point occurs where the derivative of the function, \(f'(x)\), is equal to zero. We start by setting \(f'(x)\) to zero and solving for \(x\): \[4x^3 - 9x = 0\].Factoring the expression, we get:\[x(4x^2 - 9) = 0\].This results in two solutions: \(x = 0\) and solving \(4x^2 - 9 = 0\). For \(4x^2 - 9 = 0\), add 9 to both sides and divide by 4 to get:\[x^2 = \frac{9}{4}\].Taking square roots gives \(x = \frac{3}{2}\) and \(x = -\frac{3}{2}\). Thus, the critical points are \(x = 0, x = \frac{3}{2}, x = -\frac{3}{2}\).

Step 2: Determine Nature of Critical Points

To determine whether each critical point is a relative maximum, minimum, or neither, use the second derivative test. Calculate the second derivative, \(f''(x)\):\[f''(x) = (4x^3 - 9x)' = 12x^2 - 9\].- **At** \(x = 0\): \[f''(0) = 12(0)^2 - 9 = -9\]. Since \(f''(0) < 0\), there is a relative maximum at \(x = 0\).- **At** \(x = \frac{3}{2}\): \[f''\left(\frac{3}{2}\right) = 12\left(\frac{3}{2}\right)^2 - 9 = 27 \]. Since \(f''\left(\frac{3}{2}\right) > 0\), there is a relative minimum at \(x = \frac{3}{2}\).- **At** \(x = -\frac{3}{2}\): \[f''\left(-\frac{3}{2}\right) = 12\left(-\frac{3}{2}\right)^2 - 9 = 27\]. Since \(f''\left(-\frac{3}{2}\right) > 0\), there is a relative minimum at \(x = -\frac{3}{2}\).

Key Concepts

Second Derivative Test

In calculus, the second derivative test helps determine if a function's critical points are relative maxima, minima, or neither.
It's a simple tool based on the second derivative of a function. If you have a function whose first derivative, denoted as \(f'(x)\), is zero at some point \(x = c\), this point is known as a critical point.
Here's how to use the test:

• Find the second derivative of the function, \(f''(x)\).
• Evaluate \(f''(x)\) at the critical point \(c\).
• If \(f''(c) > 0\), the function has a relative minimum at \(c\).
• If \(f''(c) < 0\), the function has a relative maximum at \(c\).
• If \(f''(c) = 0\), the test is inconclusive; you cannot determine the behavior based solely on the second derivative.

The test is a great way to quickly analyze the nature of a critical point without sketching the function graph.
It relies on the concavity of the function at those points.

Relative Maximum

A relative maximum is a point where the function reaches a peak value relative to nearby points.
This means that at this specific point, the function's value is higher than the values of the function immediately around it.
When employing the second derivative test as previously discussed, look for where the second derivative, \(f''(x)\), is less than zero.
You can think about it this way: the function curves downward at these points, creating a local high point or peak.
It's like standing on a small hill; the top of the hill (your relative maximum) is higher than the ground on any side.
Remember, a relative maximum doesn't mean it's the highest point on the entire function, just higher compared to others nearby.
This local approach is what makes it 'relative'.

Relative Minimum

A relative minimum occurs when a function reaches a low point compared to nearby values.
This means that the function's value at a relative minimum will be lower than its values in the immediate vicinity.
In the context of the second derivative test, a point is identified as a relative minimum if the second derivative \(f''(x)\) at that point is greater than zero.
This scenario implies that the graph of the function is curving upwards, forming a trough or valley.
Imagine standing at the bottom of a small valley; here, the ground directly around you is higher, signifying a local low point, or relative minimum.
It's important to understand that this doesn't indicate the absolute lowest point of the function—just the lowest within its vicinity.
That's why it's called a 'relative' minimum.

Finding Derivatives

Finding derivatives is a fundamental calculus operation that provides insights into how a function changes at any point.
Derivatives essentially measure the rate of change or the slope of a function at a given point, and it's especially useful in finding critical points.
Here's a basic way to find a derivative of a polynomial function:

• For a term like \(ax^n\), the derivative is \(n \cdot ax^{n-1}\).
• Apply this rule to each term of the function separately.
• Combine the results to get the derivative of the entire function.

Once you have the first derivative, set it to zero to find the critical points where the rate of change halts, potentially marking maxima, minima, or points of inflection.
Understanding how to find derivatives efficiently allows you to better analyze the behavior of functions and solve a wide array of real-world problems.
With practice, finding derivatives becomes a routine part of analyzing and interpreting functions.