Problem 51 - Calculus with Concepts in Calculus

Question

Find a point \(c\) such that \(f^{\prime}(c)=f^{\prime \prime}(c)=0\) Using the derivative of \(f\), show that \(f\) does not have a relative extreme value at \(c\). $$ f(x)=x^{5}-x^{3} $$

Step-by-Step Solution

Answer

The point is \( x = 0 \), and there is no relative extreme value at this point.

Step 1: Find the First Derivative

First, we need to find the first derivative of the function \( f(x) = x^5 - x^3 \). Using the power rule, the derivative is:\[f'(x) = 5x^4 - 3x^2\]

Step 2: Set First Derivative to Zero

To find critical points where the first derivative is zero, set:\[5x^4 - 3x^2 = 0\]Factor out \(x^2\):\[x^2(5x^2 - 3) = 0\]This gives solutions \(x^2 = 0\) or \(5x^2 - 3 = 0\). So, \(x = 0\) or \(x = \pm\sqrt{\frac{3}{5}}\).

Step 3: Find the Second Derivative

Next, find the second derivative of \( f(x) = x^5 - x^3 \) using the first derivative we found in Step 1:\[f''(x) = \frac{d}{dx}(5x^4 - 3x^2) = 20x^3 - 6x\]

Step 4: Identify Points where Both Derivatives are Zero

We need both \( f'(c) = 0 \) and \( f''(c) = 0 \). We already have \( f'(x) = 0 \) at \(x = 0\), and substitute into the second derivative:\[f''(0) = 20(0)^3 - 6(0) = 0\]Thus, \(x = 0\) is the point where both derivatives are zero.

Step 5: Analyze the Behavior of the Function at \( c \)

To determine if \(f\) has a relative extreme at \(x=0\), evaluate higher-order derivatives or use a sign analysis around the point. Here, for simplicity, check the sign change of the first derivative's neighboring points around \( x=0 \).Pick points like \(-0.1\) and \(+0.1\), and substitute into \(f'(x)\): \( f'(-0.1) = 5(-0.1)^4 - 3(-0.1)^2 = 0.005 - 0.03 = -0.025 < 0\) \( f'(0.1) = 5(0.1)^4 - 3(0.1)^2 = 0.005 - 0.03 = -0.025 < 0\) There is no sign change, indicating \(f(x)\) does not have a relative extreme at \( x=0 \).

Key Concepts

Derivatives

A derivative shows the rate at which a function is changing at any given point. It tells us how much a function's output changes per small change in its input, essentially describing the slope of the tangent line at any point on the function.

• To find the derivative of a polynomial like \(f(x) = x^5 - x^3\), we use the power rule.
• The power rule states that for each term \(ax^n\), the derivative is \(n \cdot ax^{n-1}\).

In this case:- The derivative of \(x^5\) is \(5x^4\).- The derivative of \(-x^3\) is \(-3x^2\).
Thus, the first derivative of the function is \(f'(x) = 5x^4 - 3x^2\). This expression helps to identify critical points and the behavior of the function.

Critical Points

Critical points are values of \(x\) where the first derivative \(f'(x)\) is zero or undefined. These points can signal potential local maximums, minimums, or points of inflection where the function changes concavity or does not change it.

• For \(f(x) = x^5 - x^3\), we set the derivative \(5x^4 - 3x^2 = 0\) to find critical points.
• Factoring gives us \(x^2(5x^2 - 3) = 0\).

This equation reveals the critical points:- \(x = 0\) when \(x^2 = 0\)- \(x = \pm\sqrt{\frac{3}{5}}\) when \(5x^2 - 3 = 0\)
These are the points where the function's slope is zero.

Extreme Values

Extreme values indicate the highest or lowest points in a section of the curve, known as the local maximums or minimums.

• To determine if a critical point is a local extreme value, further analysis is needed beyond finding where the derivative equals zero.

At the critical point \(x = 0\), both the first and second derivatives equal zero, \(f'(c) = f''(c) = 0\).
However,- There is no sign change in the first derivative around \(x=0\) indicating it's not a local extreme.- The same is true for points \(-0.1\) and \(0.1\), further confirming no extreme value at \(x=0\). This analysis shows a flat point but not an extreme one.

Second Derivative Test

The second derivative test helps decide if a critical point is a local maximum, minimum, or neither. It uses the concavity at the critical point.

• The second derivative of \(f(x) = x^5 - x^3\) is \(f''(x) = 20x^3 - 6x\).

To apply the test,- Find \(f''(c)\).- If \(f''(c) > 0\), the point is a local minimum.- If \(f''(c) < 0\), the point is a local maximum.- If \(f''(c) = 0\), the test is inconclusive.
At \(x = 0\), both \(f'(0)\) and \(f''(0)\) are zero. Therefore, the test is inconclusive and does not indicate a local extreme value at this point. Further analysis or higher-order derivatives might be needed to fully understand the curve's behavior at such critical points.