The graph of $$ f(x)=x^{3}+4 x^{2} $$ is concave downward on which interval(s)? $$ \begin{array}{llll} ext { A. }\left(-\frac{3}{4}, \infty ight) & ext { B. }(0, \infty) & ext { C. }\left(-\infty,-\frac{4}{3} ight) & ext { D. }\left(-\frac{4}{3}, \infty ight)\end{array} $$ E. $$ \left(-\infty,-\frac{3}{4} ight) $$

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The graph of $$ f(x)=x^{3}+4 x^{2} $$ is concave downward on which interval(s)? $$ \begin{array}{llll}	ext { A. }\left(-\frac{3}{4}, \inftyight) & 	ext { B. }(0, \infty) & 	ext { C. }\left(-\infty,-\frac{4}{3}ight) & 	ext { D. }\left(-\frac{4}{3}, \inftyight)\end{array} $$ E. $$ \left(-\infty,-\frac{3}{4}ight) $$

UpStudy AI · Accepted Answer

To determine where the graph of $$ f(x) = x^3 + 4x^2 $$ is concave downward, we need to analyze the second derivative of the function.

1. **First Derivative:**
   $$
   f'(x) = 3x^2 + 8x
   $$

2. **Second Derivative:**
   $$
   f''(x) = 6x + 8
   $$

3. **Determine Concavity:**
   - The graph is **concave downward** where $$ f''(x) < 0 $$.
   $$
   6x + 8 < 0
   $$
   $$
   6x < -8
   $$
   $$
   x < -\frac{4}{3}
   $$

Therefore, the function is concave downward on the interval $$ \left(-\infty, -\frac{4}{3}\right) $$.

**Answer:**  
C. $$\left(-\infty,-\frac{4}{3}\right)$$
The graph of $$ f(x) = x^3 + 4x^2 $$ is concave downward on the interval $$ \left(-\infty, -\frac{4}{3}\right) $$.

The graph of is concave downward on which interval(s)?
E.

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