Questions: Consider the function (f(x)) below. Over what open interval(s) is the function increasing and concave up? Give your answer in interval notation. [f(x)=fracx^44-frac2 x^33-2 x^2+8 x-8] Enter (varnothing) if the interval does not exist.

Solution Steps

To determine where the function \( f(x) \) is increasing, we first compute its first derivative: \[ f'(x) = \frac{d}{dx}\left(\frac{x^{4}}{4}-\frac{2x^{3}}{3}-2x^{2}+8x-8\right). \] Using the power rule, we get: \[ f'(x) = x^{3} - 2x^{2} - 4x + 8. \]

To find the critical points, solve \( f'(x) = 0 \): \[ x^{3} - 2x^{2} - 4x + 8 = 0. \] Factoring the equation: \[ (x - 2)(x^{2} - 4) = 0. \] This gives the critical points: \[ x = 2, \quad x = 2, \quad x = -2. \] Thus, the critical points are \( x = -2 \) and \( x = 2 \).

To determine where \( f(x) \) is increasing, analyze the sign of \( f'(x) \) in the intervals determined by the critical points:

• For \( x < -2 \), choose \( x = -3 \): \[ f'(-3) = (-3)^{3} - 2(-3)^{2} - 4(-3) + 8 = -27 - 18 + 12 + 8 = -25 < 0. \] Thus, \( f(x) \) is decreasing on \( (-\infty, -2) \).

• For \( -2 < x < 2 \), choose \( x = 0 \): \[ f'(0) = 0^{3} - 2(0)^{2} - 4(0) + 8 = 8 > 0. \] Thus, \( f(x) \) is increasing on \( (-2, 2) \).

• For \( x > 2 \), choose \( x = 3 \): \[ f'(3) = 3^{3} - 2(3)^{2} - 4(3) + 8 = 27 - 18 - 12 + 8 = 5 > 0. \] Thus, \( f(x) \) is increasing on \( (2, \infty) \).

To determine where the function is concave up, compute the second derivative: \[ f''(x) = \frac{d}{dx}\left(x^{3} - 2x^{2} - 4x + 8\right) = 3x^{2} - 4x - 4. \]

Solve \( f''(x) > 0 \): \[ 3x^{2} - 4x - 4 > 0. \] Find the roots of \( 3x^{2} - 4x - 4 = 0 \): \[ x = \frac{4 \pm \sqrt{(-4)^{2} - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6}. \] Thus, the roots are: \[ x = 2, \quad x = -\frac{2}{3}. \] Analyze the sign of \( f''(x) \) in the intervals determined by the roots:

• For \( x < -\frac{2}{3} \), choose \( x = -1 \): \[ f''(-1) = 3(-1)^{2} - 4(-1) - 4 = 3 + 4 - 4 = 3 > 0. \] Thus, \( f(x) \) is concave up on \( (-\infty, -\frac{2}{3}) \).

• For \( -\frac{2}{3} < x < 2 \), choose \( x = 0 \): \[ f''(0) = 3(0)^{2} - 4(0) - 4 = -4 < 0. \] Thus, \( f(x) \) is concave down on \( (-\frac{2}{3}, 2) \).

• For \( x > 2 \), choose \( x = 3 \): \[ f''(3) = 3(3)^{2} - 4(3) - 4 = 27 - 12 - 4 = 11 > 0. \] Thus, \( f(x) \) is concave up on \( (2, \infty) \).

The function \( f(x) \) is increasing on \( (-2, 2) \) and \( (2, \infty) \), and concave up on \( (-\infty, -\frac{2}{3}) \) and \( (2, \infty) \). The intersection of these intervals is: \[ (2, \infty). \]

Final Answer