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 inflection points. $$ f(x)=x^{16}+3 x^{2} $$

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UpStudy AI · Accepted Answer

**Step 1. Compute the first derivative**  
We start with  
$$
f(x) = x^{16} + 3x^2.
$$  
The first derivative is  
$$
f'(x) = 16x^{15} + 6x.
$$

**Step 2. Compute the second derivative**  
Differentiate $$ f'(x) $$ to obtain  
$$
f''(x) = 240x^{14} + 6.
$$

**Step 3. Determine the concavity**  
The concavity of the graph is determined by the sign of $$ f''(x) $$. Notice that for any real number $$ x $$,  
$$
x^{14} \geq 0,
$$  
so  
$$
240x^{14} \geq 0.
$$  
Thus,  
$$
f''(x) = 240x^{14} + 6 \geq 6.
$$  
Since $$ 6 > 0 $$, it follows that  
$$
f''(x) > 0 \quad \text{for all } x.
$$

**Step 4. Conclude Concavity and Inflection Points**  
- The graph is concave upward on the interval $$(-\infty, \infty)$$ because $$ f''(x) > 0 $$ for all $$ x $$.  
- There are no intervals where the graph is concave downward.  
- An inflection point occurs where $$ f''(x) $$ changes sign. Since $$ f''(x) $$ is always positive and never equals zero, there are no inflection points.

**Final Answer:**  
- Concave upward on: $$(-\infty, \infty)$$  
- Concave downward on: *None*  
- Inflection points: *None*
The graph of $$ f(x) = x^{16} + 3x^2 $$ is concave upward everywhere, with no intervals where it's concave downward or inflection points.

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 inflection points. \( f(x)=x^{16}+3 x^{2} \)

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