Problem 5 - Calculus - AP Edition

Question

(a) Show that both of the functions \(f(x)=(x-1)^{4}\) and \(g(x)=x^{3}-3 x^{2}+3 x-2\) have stationary points at \(x=1\). (b) What does the second derivative test tell you about the nature of these stationary points? (c) What does the first derivative test tell you about the nature of these stationary points?

Step-by-Step Solution

Answer

Both have stationary points at \( x=1 \); \( f(x) \) has a local minimum, \( g(x) \) a point of inflection.

Step 1: Find the derivative of the function f(x)

To find the stationary points, we first need to find the derivative of \( f(x) = (x-1)^4 \). Using the chain rule, the derivative is: \[ f'(x) = 4(x-1)^3 \].

Step 2: Find stationary points for f(x)

Stationary points occur when the derivative equals zero. Set \( f'(x) = 0 \): \[ 4(x-1)^3 = 0 \] Solving for \( x \), we get \( x-1=0 \), thus \( x=1 \).

Step 3: Find the derivative of the function g(x)

Now, let's find the derivative of \( g(x) = x^3 - 3x^2 + 3x - 2 \). The derivative is: \[ g'(x) = 3x^2 - 6x + 3 \].

Step 4: Find stationary points for g(x)

Set the derivative \( g'(x) = 0 \) to find stationary points: \[ 3x^2 - 6x + 3 = 0 \] Factor the equation: \( 3(x-1)^2 = 0 \) which gives \( x = 1 \).

Step 5: Apply the second derivative test for f(x)

Calculate the second derivative \( f''(x) \): \[ f''(x) = 12(x-1)^2 \] Evaluate \( f''(1) \): \[ f''(1) = 12(1-1)^2 = 0 \] The test is inconclusive since \( f''(1) = 0 \).

Step 6: Apply the second derivative test for g(x)

Calculate the second derivative \( g''(x) \): \[ g''(x) = 6x - 6 \] Evaluate \( g''(1) \): \[ g''(1) = 6(1) - 6 = 0 \] The test is inconclusive since \( g''(1) = 0 \).

Step 7: Examine f(x) with the first derivative test

Evaluate \( f'(x) = 4(x-1)^3 \) around \( x=1 \). To the left of \( x=1 \), \( f'(x) < 0 \) and to the right, \( f'(x) > 0 \). This indicates \( x=1 \) is a local minimum.

Step 8: Examine g(x) with the first derivative test

Evaluate \( g'(x) = 3(x-1)^2 \) around \( x=1 \). Since \( g'(x) = 0 \) at \( x=1 \) and does not change sign, \( x=1 \) is a point of inflection.

Key Concepts

Second Derivative Test

The second derivative test is a helpful tool in finding out the nature of stationary points found on a function. But what is it exactly? When you find a derivative and then a second derivative, you can use this test to determine whether a stationary point (where the first derivative is zero) is a local maximum, a local minimum, or inconclusive. Here’s how it works:

• If the second derivative at the stationary point is positive, the function has a local minimum at that point. Imagine a "U-shape" (like a smile).
• If it is negative, the function exhibits a local maximum (an upside-down "U-shape" like a frown).
• However, if the second derivative is zero, similar to our original exercise with both functions at \( x=1 \), the test doesn't provide a clear answer, making it inconclusive.

This specific test didn't tell us everything we needed to know about functions \( f(x) = (x-1)^4 \) and \( g(x) = x^3 - 3x^2 + 3x - 2 \). Both had a second derivative equal to zero at the stationary point \( x = 1 \). That's why we had to turn to another tool: the first derivative test.

First Derivative Test

The first derivative test helps us further analyze stationary points, particularly when the second derivative test is inconclusive. Think of it as examining changes in the function's slope.Here’s the basic idea:

• Analyze how the first derivative changes sign around the stationary point.
• If you move slightly left and then slightly right of the stationary point, you can check the sign (negative or positive) of the first derivative.
• When the derivative moves from negative to positive, you have a local minimum. If it shifts from positive to negative, you’ve found a local maximum.
• If there’s no change in sign, as with \( g(x) \) at \( x=1 \), it’s likely a point of inflection. This means the curve has a change in concavity rather than a peak or trough.

In our case, for \( f(x) \), the first derivative told us there was a local minimum because it changed from negative to positive around \( x=1 \). However, for \( g(x) \), since the sign didn’t change, we learned that \( x=1 \) wasn't a maximum or minimum, but a point of inflection.

Derivatives in Calculus

Derivatives are a fundamental concept in calculus and play a significant role in analyzing the behavior of functions. But let’s break it down: a derivative reveals the rate at which a function changes and is akin to the slope of a function at a particular point. Key points about derivatives include:

• The first derivative of a function \( f(x) \) represents the instantaneous rate of change or slope. For instance, when we differentiated \( f(x) = (x-1)^4 \), we found that \( f'(x) = 4(x-1)^3 \). This helped identify stationary points where \( f'(x) = 0 \).
• The second derivative, represented as \( f''(x) \), provides information about the curvature of the function. In our exercise, it wasn't definitive for determining the nature of the stationary points for both functions at \( x=1 \).

Through derivatives, we gain deeper insights into how functions behave, allowing us to solve practical problems such as finding maximum profits, minimizing errors, or analyzing the efficiency of systems. In calculus, mastering derivatives opens up a world of understanding the dynamics of change.