Problem 11 - Applied Calculus

Question

Find the inflection points of \(f(x)=x^{4}+x^{3}-3 x^{2}+2\).

Step-by-Step Solution

Answer

The inflection points are at \( x = -1 \) and \( x = \frac{1}{2} \).

Step 1: Find the Second Derivative

To find the inflection points, we first need the second derivative of the function. Start by computing the first derivative of the function:Given: \( f(x) = x^4 + x^3 - 3x^2 + 2 \)First Derivative: \( f'(x) = 4x^3 + 3x^2 - 6x \)Now, calculate the second derivative:Second Derivative: \( f''(x) = 12x^2 + 6x - 6 \)

Step 2: Set the Second Derivative to Zero

The inflection points occur where the second derivative is zero. Solve the equation:\( f''(x) = 12x^2 + 6x - 6 = 0 \)Divide the entire equation by 6 for simplification:\( 2x^2 + x - 1 = 0 \)

Step 3: Solve the Quadratic Equation

Solve \( 2x^2 + x - 1 = 0 \) using the quadratic formula:The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).Here, \( a = 2 \), \( b = 1 \), and \( c = -1 \).Calculate the discriminant: \( b^2 - 4ac = 1 + 8 = 9 \).The solutions are:\( x = \frac{-1 \pm 3}{4} \).Thus, \( x = \frac{1}{2} \) and \( x = -1 \).

Step 4: Verify Change in Concavity

To confirm that these are inflection points, test the values for a change in concavity. Choose test points around \( x = -1 \) and \( x = \frac{1}{2} \).For \( x = -1 \):- Test point \( x = -1.5 \): \( f''(-1.5) = 12(-1.5)^2 + 6(-1.5) - 6 = 4.5 > 0 \)- Test point \( x = -0.5 \): \( f''(-0.5) = 12(-0.5)^2 + 6(-0.5) - 6 = -4.5 < 0 \)For \( x = \frac{1}{2} \):- Test point \( x = 0 \): \( f''(0) = 12(0)^2 + 6(0) - 6 = -6 < 0 \)- Test point \( x = 1 \): \( f''(1) = 12(1)^2 + 6(1) - 6 = 12 > 0 \)Both points \( x = -1 \) and \( x = \frac{1}{2} \) have a change in concavity, confirming they are inflection points.

Key Concepts

Second Derivative

The journey to finding inflection points begins by examining the second derivative of a function. An inflection point is a point on a curve where the concavity changes. To determine this, we first find the second derivative of the function.
For a given function, like \( f(x) = x^4 + x^3 - 3x^2 + 2 \), we start by finding its first derivative. This is achieved by applying basic differentiation rules to each term:

• The term \( x^4 \) differentiates to \( 4x^3 \).
• The term \( x^3 \) becomes \( 3x^2 \).
• The term \( -3x^2 \) turns into \( -6x \).

Thus, the first derivative is \( f'(x) = 4x^3 + 3x^2 - 6x \).
Next, we differentiate this result to get the second derivative:

• \( 4x^3 \) becomes \( 12x^2 \).
• \( 3x^2 \) shifts to \( 6x \).
• \( -6x \) differentiates to \( -6 \).

This culminates in the second derivative: \( f''(x) = 12x^2 + 6x - 6 \).
We focus on where this expression equals zero since these points could signal a change in concavity, indicating potential inflection points.

Quadratic Formula

Once the second derivative is set to zero, the next step is to solve the resulting quadratic equation. In this case, the equation \( 12x^2 + 6x - 6 = 0 \) simplifies to \( 2x^2 + x - 1 = 0 \) by dividing all terms by 6.
Solving a quadratic equation typically involves factoring, completing the square, or using the quadratic formula. Here, the quadratic formula is ideal due to its general applicability:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For \( 2x^2 + x - 1 = 0 \), let \( a = 2 \), \( b = 1 \), and \( c = -1 \). Plugging these values into the formula gives:

• \( b^2 - 4ac = 1 - 4(2)(-1) = 9 \)
• The solutions are \( x = \frac{-1 \pm 3}{4} \)

This means our \( x \) values are \( \frac{1}{2} \) and \( -1 \). These values are potential points of inflection that require verification of concavity change.

Change in Concavity

The confirmation of an inflection point involves checking for a change in concavity at the values obtained from solving the second derivative equation. A change in concavity occurs when the second derivative changes sign around the point.
For our function, we must test points around \( x = -1 \) and \( x = \frac{1}{2} \):

• **Around \( x = -1 \):**
  - At \( x = -1.5 \), the second derivative is \( f''(-1.5) = 4.5 \) (positive concavity).
  - At \( x = -0.5 \), \( f''(-0.5) = -4.5 \) (negative concavity).


• **Around \( x = \frac{1}{2} \):**
  - At \( x = 0 \), \( f''(0) = -6 \) (negative concavity).
  - At \( x = 1 \), \( f''(1) = 12 \) (positive concavity).

These changes from positive to negative and vice versa confirm that \( x = -1 \) and \( x = \frac{1}{2} \) are indeed inflection points. Recognizing this change is crucial to understanding the shaping of the function's graph.