$g(x)=x^4+4x^3-48x^2+6$



On which intervals is the graph of $g$ concave down?



Choose 1 answer:

A(-4, 2) only
B$(-\infty,-2)$ and $(4, \infty)$
C$(-2, \infty)$ only
D$(-\infty,-4)$ and $(2, \infty)$

Verified Answer :

  • Compute the second derivative of the function: \(g''(x) = 12x^2 + 24x - 96\).
  • Solve for the roots of the second derivative, which are \(x = -4\) and \(x = 2\).
  • Test the intervals \((-\infty, -4)\), \((-4, 2)\), and \((2, \infty)\) by evaluating the sign of \(g''(x)\) in each.
  • Conclude that the function is concave down on the interval \((-4, 2)\), which corresponds to option \(\boxed{(A)}\).

Explanation

  1. Problem Analysis and Concavity Condition

To determine the intervals where the graph of the function \(g(x) = x^4 + 4x^3 - 48x^2 + 6\) is concave down, we must analyze the sign of its second derivative, \(g''(x)\). A function is concave down on intervals where its second derivative is negative.

  1. Computing the Derivatives

First, we calculate the first derivative of \(g(x)\):
\[g'(x) = \frac{d}{dx}(x^4 + 4x^3 - 48x^2 + 6) = 4x^3 + 12x^2 - 96x\]
Next, we calculate the second derivative of \(g(x)\) by differentiating \(g'(x)\):
\[g''(x) = \frac{d}{dx}(4x^3 + 12x^2 - 96x) = 12x^2 + 24x - 96\]

  1. Finding Critical Points of the Second Derivative

To find the points where the concavity might change, we set the second derivative equal to zero:
\[12x^2 + 24x - 96 = 0\]
Dividing the entire equation by 12 simplifies it to:
\[x^2 + 2x - 8 = 0\]
Factoring the quadratic equation gives:
\[(x + 4)(x - 2) = 0\]
Thus, the roots are \(x = -4\) and \(x = 2\). These points divide the real number line into three intervals: \((-\infty, -4)\), \((-4, 2)\), and \((2, \infty)\).

  1. Determining Concavity Intervals

Now, we test each interval to determine the sign of \(g''(x)\):

  • For the interval \((-\infty, -4)\), choose \(x = -5\):

\(g''(-5) = 12(-5)^2 + 24(-5) - 96 = 12(25) - 120 - 96 = 300 - 216 = 84 > 0\). (Concave up)

  • For the interval \((-4, 2)\), choose \(x = 0\):

\(g''(0) = 12(0)^2 + 24(0) - 96 = -96 < 0\). (Concave down)

  • For the interval \((2, \infty)\), choose \(x = 3\):

\(g''(3) = 12(3)^2 + 24(3) - 96 = 108 + 72 - 96 = 84 > 0\). (Concave up)
The second derivative is negative only on the interval \((-4, 2)\).

  1. Final Answer

Based on our analysis, the graph of \(g(x)\) is concave down on the interval \((-4, 2)\). Comparing this to the given choices, this matches option (A). The final answer is \(\boxed{(A)}\).

Examples

In engineering, understanding the concavity of a curve is essential for structural stability. For instance, if you are designing the curvature of a bridge or a roller coaster track, identifying where the structure is "concave down" (arching) versus "concave up" (dipping) helps in calculating the stress distribution and ensuring the structure maintains its integrity under load. This calculus-based analysis is the fundamental step in determining such geometric properties.