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. $$ f(x)=x^{3}-15 x^{2}+2 x+3 $$ For what interval(s) of $$ x $$ is the graph of $$ f $$ concave upward? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. Type your answer in interval notation. Type an exact answer. Use a comma to separate answers as needed.) B. The graph is never concave upward. For what interval(s) of $$ x $$ is the graph of $$ f $$ concave downward? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. B. The graph is never concave downward. Determine the $$ x, y $$ coordinates of the inflection point of the graph of $$ f $$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. Type an ordered pair.) (Typerval notation. Type an exact answer. Use a comma to separate answers as needed.) (T)

Question

UpStudy AI · Accepted Answer

The first step is to find the second derivative of $$ f(x) $$.

1. Compute the first derivative:
$$
f(x) = x^3 - 15x^2 + 2x + 3 \quad \Longrightarrow \quad f'(x) = 3x^2 - 30x + 2.
$$

2. Compute the second derivative:
$$
f''(x) = \frac{d}{dx}\left(3x^2 - 30x + 2\right) = 6x - 30.
$$

3. Determine the intervals of concavity by finding where $$ f''(x) $$ is positive or negative.
   - Set $$ f''(x) = 0 $$:
   $$
   6x - 30 = 0 \quad \Longrightarrow \quad x = 5.
   $$
   This divides the real line into two intervals: $$ (-\infty, 5) $$ and $$ (5, \infty) $$.

4. Test each interval:
   - For $$ x < 5 $$ (choose $$ x = 0 $$):
   $$
   f''(0) = 6(0) - 30 = -30 \quad (\text{negative}),
   $$
   so the graph is concave downward on $$ (-\infty, 5) $$.

- For $$ x > 5 $$ (choose $$ x = 6 $$):
   $$
   f''(6) = 6(6) - 30 = 36 - 30 = 6 \quad (\text{positive}),
   $$
   so the graph is concave upward on $$ (5, \infty) $$.

5. Find the inflection point where the concavity changes (at $$ x = 5 $$):
   $$
   f(5) = 5^3 - 15(5^2) + 2(5) + 3 = 125 - 375 + 10 + 3 = -237.
   $$
   Thus, the inflection point is $$ (5, -237) $$.

The answers are:

- The graph is concave upward on: $$ (5, \infty) $$.
- The graph is concave downward on: $$ (-\infty, 5) $$.
- The inflection point is: $$ (5, -237) $$.
The graph of $$ f(x) = x^3 - 15x^2 + 2x + 3 $$ is concave upward on the interval $$ (5, \infty) $$, concave downward on $$ (-\infty, 5) $$, and has an inflection point at $$ (5, -237) $$.

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Calculus Questions