Problem 5 - Calculus A New Horizon

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 functions have stationary points at \(x=1\). The second derivative test is inconclusive; the first derivative test shows \(f(x)\) has a local minimum, \(g(x)\) a point of inflection at \(x=1\).

Step 1: Find the First Derivative of f(x)

To find the stationary points of \(f(x)=(x-1)^4\), first find its first derivative. Using the power rule, \(f'(x) = 4(x-1)^3\).

Step 2: Solve f'(x) = 0 for Stationary Points

Set \(f'(x) = 4(x-1)^3 = 0\) and solve for \(x\). Solving, we get: \(4(x-1)^3 = 0\). Dividing by 4 and taking the cube root gives \((x-1) = 0\), thus \(x = 1\) is a stationary point.

Step 3: Find the First Derivative of g(x)

Differentiate \(g(x) = x^3 - 3x^2 + 3x - 2\) with respect to \(x\). Using the power rule, \(g'(x) = 3x^2 - 6x + 3\).

Step 4: Solve g'(x) = 0 for Stationary Points

Set \(g'(x) = 3x^2 - 6x + 3 = 0\) and solve for \(x\): Dividing through by 3 gives \(x^2 - 2x + 1 = 0\). Factoring, we find \((x-1)^2 = 0\), so \(x = 1\) is a stationary point.

Step 5: Calculate Second Derivative of f(x) for Nature of Stationary Point

To use the second derivative test on \(f(x)\), find \(f''(x)\). Differentiating \(f'(x) = 4(x-1)^3\) gives \(f''(x) = 12(x-1)^2\). Evaluating at \(x=1\), \(f''(1) = 12 \times 0^2 = 0\), so the test is inconclusive.

Step 6: Calculate Second Derivative of g(x) for Nature of Stationary Point

For \(g(x)\), find \(g''(x)\) by differentiating \(g'(x) = 3x^2 - 6x + 3\).This yields \(g''(x) = 6x - 6\). Evaluating at \(x=1\), \(g''(1) = 6 \times 1 - 6 = 0\), so the test is inconclusive.

Step 7: Use First Derivative Test for f(x) to Determine Stationary Point Nature

Consider intervals around \(x=1\) for \(f'(x) = 4(x-1)^3\). Before \(x=1\) (say \(x=0\)), \(f'(x)\) is negative; after \(x=1\) (say \(x=2\)), \(f'(x)\) is positive. Transition from negative to positive suggests a local minimum at \(x=1\).

Step 8: Use First Derivative Test for g(x) to Determine Stationary Point Nature

For \(g'(x) = 3(x-1)^2\), note that \(g'(x) = 0\) on both sides of \(x=1\), indicating no change in sign. This suggests the point \(x=1\) is a stationary point of inflection.

Key Concepts

First Derivative Test

The first derivative test is valuable for determining the nature of stationary points. A stationary point occurs where the first derivative of a function equals zero, indicating no slope at that point. This test involves examining the sign changes of the first derivative before and after the stationary point.

• If the first derivative changes from negative to positive as we cross the point, the slope transitions from decreasing to increasing, indicating a local minimum.
• If the first derivative changes from positive to negative, the function is increasing then decreasing, suggesting a local maximum.
• If there is no sign change, we might have a stationary point of inflection, where the curve doesn't exhibit a peak or trough effect.

For the functions provided, when examining \(f(x) = (x-1)^4\):

• The transition from a negative to a positive derivative suggests a local minimum at \(x=1\).

For \(g(x) = x^3-3x^2+3x-2\):

• With the first derivative zero throughout, this suggests a stationary point of inflection at \(x=1\), showing no curvature change around this point.

Understanding how the slope behaves around stationary points is crucial in sketching and analyzing function behaviors.

Second Derivative Test

The second derivative test is another technique to further evaluate the nature of stationary points. It provides insight by assessing the concavity or "curviness" of the function at the stationary point by applying the second derivative.

• If the second derivative is positive at the point, the curve is concave up, indicating a local minimum.
• If negative, the curve is concave down, showing a local maximum.
• If the second derivative himself is zero, the test is inconclusive regarding whether the point is a maximum, minimum, or an inflection point.

For the provided functions:

• For \(f(x) = (x-1)^4\), evaluating the second derivative \(f''(1) = 0\), renders the second derivative test inconclusive.
• Similarly, for \(g(x) = x^3-3x^2+3x-2\), the second derivative at \(x=1\) is also zero, again leaving the test inconclusive.

When the second derivative fails to conclusively identify the nature, the first derivative test or other methods may need to be used to understand the behavior better.

Calculus Differentiation

Differentiation is a key concept in calculus, utilized to find the derivative of a function. The derivative represents the rate at which the function value changes as its input changes. It is fundamental in identifying extreme values in functions, which are critical for many applications in mathematics, physics, and engineering.
Differentiation involves several rules that simplify calculating derivatives:

• The power rule states that for \(f(x) = x^n\), \(f'(x) = nx^{n-1}\).
• The sum rule allows derivatives of sums to be handled term by term, i.e., \((u+v)' = u' + v'\).
• The product and quotient rules handle products and divisions of functions.

For the functions in the exercise:

• \(f(x) = (x-1)^4\) applied the power rule to find \(f'(x) = 4(x-1)^3\).
• \(g(x) = x^3-3x^2+3x-2\) used the power rule separately across terms to arrive at \(g'(x) = 3x^2 - 6x + 3\).

Proficiency in these differentiation rules helps in performing the necessary calculations to solve real-world and theoretical calculus problems efficiently.