Verified Answer :

  • Compute the first and second derivatives of the function to be \(h'(x) = 4x^3 - 4x\) and \(h''(x) = 12x^2 - 4\).
  • Identify candidates for inflection points by setting \(h''(x) = 0\), which results in \(x = \pm \frac{\sqrt{3}}{3}\).
  • Verify the change in concavity by testing the sign of \(h''(x)\) in the intervals around these points.
  • Calculate the corresponding y-values to find the exact coordinates of the inflection points.
  • The inflection points are \(\boxed{\left(-\frac{\sqrt{3}}{3}, \frac{49}{9}\right), \left(\frac{\sqrt{3}}{3}, \frac{49}{9}\right)}\).

Explanation

  1. Problem Analysis and Strategy

The function provided is \(h(x) = x^4 - 2x^2 + 6\). An inflection point occurs where the concavity of a function changes, which corresponds to where the second derivative \(h''(x)\) changes sign. Since \(h(x)\) is a polynomial, it is continuous and differentiable everywhere. We will find its derivatives to locate these potential points.

  1. Computing Derivatives

First, we compute the first derivative using the power rule:
\[h'(x) = \frac{d}{dx}(x^4 - 2x^2 + 6) = 4x^3 - 4x\]
Next, we compute the second derivative by differentiating the first derivative:
\[h''(x) = \frac{d}{dx}(4x^3 - 4x) = 12x^2 - 4\]

  1. Finding Candidates for Inflection Points

To find the candidates for points of inflection, we set the second derivative to zero:
\[12x^2 - 4 = 0\]
\[12x^2 = 4\]
\[x^2 = \frac{4}{12} = \frac{1}{3}\]
Solving for \(x\), we get:
\[x = \pm \sqrt{\frac{1}{3}} = \pm \frac{\sqrt{3}}{3}\]
These are the critical values where the concavity might change.

  1. Verifying Concavity Changes

We test the sign of \(h''(x) = 12x^2 - 4\) in the intervals defined by \(x = -\frac{\sqrt{3}}{3}\) and \(x = \frac{\sqrt{3}}{3}\):

  1. For \(x < -\frac{\sqrt{3}}{3}\) (e.g., \(x = -1\)): \(h''(-1) = 12(-1)^2 - 4 = 8 > 0\), so the function is concave up.
  2. For \(-\frac{\sqrt{3}}{3} < x < \frac{\sqrt{3}}{3}\) (e.g., \(x = 0\)): \(h''(0) = 12(0)^2 - 4 = -4 < 0\), so the function is concave down.
  3. For \(x > \frac{\sqrt{3}}{3}\) (e.g., \(x = 1\)): \(h''(1) = 12(1)^2 - 4 = 8 > 0\), so the function is concave up.

Since the sign of \(h''(x)\) changes at both \(x = -\frac{\sqrt{3}}{3}\) and \(x = \frac{\sqrt{3}}{3}\), these are confirmed as inflection points.

  1. Evaluating Function and Final Result

Finally, we evaluate the function \(h(x) = x^4 - 2x^2 + 6\) at these x-values:
\[h\left(\pm \frac{\sqrt{3}}{3}\right) = \left(\pm \frac{\sqrt{3}}{3}\right)^4 - 2\left(\pm \frac{\sqrt{3}}{3}\right)^2 + 6 = \frac{9}{81} - 2\left(\frac{3}{9}\right) + 6 = \frac{1}{9} - \frac{2}{3} + 6 = \frac{1}{9} - \frac{6}{9} + \frac{54}{9} = \frac{49}{9}\]
Thus, the inflection points are \(\left(-\frac{\sqrt{3}}{3}, \frac{49}{9}\right)\) and \(\left(\frac{\sqrt{3}}{3}, \frac{49}{9}\right)\).

Examples

In civil engineering, determining the points of inflection on a beam's deflection curve is essential for structural analysis. When a beam is loaded, the inflection points mark the transition from downward to upward bending (or vice-versa), which dictates where the internal bending moment is zero. Engineers use this information to decide where to place reinforcement or supports to ensure the structure remains safe and balanced under stress.

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