Please answer the question providing all steps and clear writing so I can understand. Thanks.You can see as a Reference book "James Stewart Calculus 8th Edition" to see similar solutions ### Problem 8: Analyzing Inflection Points and Concavity Intervals#### Problem StatementFind the inflection points of the graph of \( f(x) = x^4 - 6x^2 \), and state the intervals on which the graph of \( f \) is concave up and the intervals on which the graph is concave down.#### ExplanationTo solve this problem, you need to follow these steps:1. **Find the Second Derivative**: Calculate the first and second derivatives of \( f(x) \).2. **Determine Inflection Points**: Set the second derivative equal to zero and solve for \( x \). These are potential inflection points.3. **Test Intervals**: Analyze the sign of the second derivative on the intervals determined by these points to establish concavity.#### Solution Steps1. **First Derivative**: \[ f'(x) = \frac{d}{dx}(x^4 - 6x^2) = 4x^3 - 12x \]2. **Second Derivative**: \[ f''(x) = \frac{d}{dx}(4x^3 - 12x) = 12x^2 - 12 \]3. **Finding Inflection Points**: Set \( f''(x) = 0 \): \[ 12x^2 - 12 = 0 \] \[ x^2 = 1 \] \[ x = \pm 1 \] Therefore, the potential inflection points are \( x = 1 \) and \( x = -1 \).4. **Test the Sign of \( f''(x) \)** on intervals \( (-\infty, -1) \), \( (-1, 1) \), and \( (1, \infty) \): - For \( x < -1 \): \[ f''(x) = 12x^2 - 12 > 0 \] So, \( f \) is concave up on \( (-\infty, -1) \). - For \( -1 < x < 1 \): \[ f''(x) = 12x^2 - 12 < 0 \] So, \( f \) is concave down on \( (-1

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Description: Please answer the question providing all steps and clear writing so I can understand. Thanks.You can see as a Reference book "James Stewart Calculus 8th Edition" to see similar solutions ### Problem 8: Analyzing Inflection Points and Concavity Interv