Questions: Find the intervals on which the graph of (f) is concave upward, the intervals on which the graph of (f) is concave downward, and the (x) coordinates of any inflection points of the graph of (f(x)). Given: [f(x)=-x^4+32 x^3-32 x+11] Determine the (x) coordinates of any inflection points of the graph of (f(x)). Select the correct choice below and, if necessary, fill in the blank to complete your choice. A. (x=) (Type an exact answer. Use a comma to separate answers as needed.) B. There are no inflection points.

Solution Steps

To determine the concavity of the function \( f(x) = x^4 - 32x^2 - 32x + 11 \), we need to find the second derivative. First, find the first derivative \( f'(x) \):

\[ f'(x) = \frac{d}{dx}(x^4 - 32x^2 - 32x + 11) = 4x^3 - 64x - 32 \]

Next, find the second derivative \( f''(x) \):

\[ f''(x) = \frac{d}{dx}(4x^3 - 64x - 32) = 12x^2 - 64 \]

Inflection points occur where the second derivative changes sign, which happens at the roots of \( f''(x) \). Set \( f''(x) = 0 \) and solve for \( x \):

\[ 12x^2 - 64 = 0 \] \[ 12x^2 = 64 \] \[ x^2 = \frac{64}{12} \] \[ x^2 = \frac{16}{3} \] \[ x = \pm \sqrt{\frac{16}{3}} \] \[ x = \pm \frac{4}{\sqrt{3}} \] \[ x = \pm \frac{4\sqrt{3}}{3} \]

To determine the concavity, test the intervals around the inflection points \( x = \pm \frac{4\sqrt{3}}{3} \). Choose test points in the intervals \( (-\infty, -\frac{4\sqrt{3}}{3}) \), \( (-\frac{4\sqrt{3}}{3}, \frac{4\sqrt{3}}{3}) \), and \( (\frac{4\sqrt{3}}{3}, \infty) \).

• For \( x \in (-\infty, -\frac{4\sqrt{3}}{3}) \), choose \( x = -3 \): \[ f''(-3) = 12(-3)^2 - 64 = 108 - 64 = 44 \] (positive, concave up)

• For \( x \in (-\frac{4\sqrt{3}}{3}, \frac{4\sqrt{3}}{3}) \), choose \( x = 0 \): \[ f''(0) = 12(0)^2 - 64 = -64 \] (negative, concave down)

• For \( x \in (\frac{4\sqrt{3}}{3}, \infty) \), choose \( x = 3 \): \[ f''(3) = 12(3)^2 - 64 = 108 - 64 = 44 \] (positive, concave up)

Final Answer