91Ó°ÊÓ

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.

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.

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.