Question

find the open intervals where the function is concave upward or concave downward. find any inflection points.
f(x)=-3x³ + 6x² + 165x - 5
where is the function concave upward and where is it concave downward? select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
o a. the function is concave upward on the interval(s) □ and concave downward on the interval(s) □.
(type your answers in interval notation. use integers or fractions for any numbers in the expressions. use a comma to separate answers as needed.)
o b. the function is concave upward on the interval(s) □. the function is never concave downward.
(type your answer in interval notation. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)
o c. the function is concave downward on the interval(s) □. the function is never concave upward.
(type your answer in interval notation. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)
o d. the function is never concave upward or downward.
find any inflection points of f. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
o a. the function has an inflection point at □.
(type an ordered pair, using integers or fractions. use a comma to separate answers as needed.)
o b. the function has no inflection points.

Explanation:

Step1: Find the first - and second - derivatives

Given \(f(x)=-3x^{3}+6x^{2}+165x - 5\).
The first derivative \(f^{\prime}(x)\) using the power rule \((x^{n})^\prime=nx^{n - 1}\):
\(f^{\prime}(x)=-9x^{2}+12x + 165\).
The second derivative \(f^{\prime\prime}(x)\):
\(f^{\prime\prime}(x)=-18x + 12\).

Step2: Find the inflection point

Set \(f^{\prime\prime}(x) = 0\).
\(-18x+12 = 0\).
Solve for \(x\):
\(18x=12\), so \(x=\frac{12}{18}=\frac{2}{3}\).

Step3: Test intervals for concavity

The inflection point \(x = \frac{2}{3}\) divides the real line into two intervals: \((-\infty,\frac{2}{3})\) and \((\frac{2}{3},\infty)\).

• For the interval \((-\infty,\frac{2}{3})\), let \(x = 0\). Then \(f^{\prime\prime}(0)=-18\times0 + 12=12>0\). So the function is concave upward on \((-\infty,\frac{2}{3})\).
• For the interval \((\frac{2}{3},\infty)\), let \(x = 1\). Then \(f^{\prime\prime}(1)=-18\times1+12=-6<0\). So the function is concave downward on \((\frac{2}{3},\infty)\).

Answer:

A. The function is concave upward on the interval \((-\infty,\frac{2}{3})\) and concave downward on the interval \((\frac{2}{3},\infty)\).
A. The function has an inflection point at \((\frac{2}{3},f(\frac{2}{3}))\). Calculate \(f(\frac{2}{3})=-3\times(\frac{2}{3})^{3}+6\times(\frac{2}{3})^{2}+165\times\frac{2}{3}-5=-3\times\frac{8}{27}+6\times\frac{4}{9}+110 - 5=-\frac{8}{9}+\frac{8}{3}+105=\frac{-8 + 24}{9}+105=\frac{16}{9}+105=\frac{16+945}{9}=\frac{961}{9}\). So the inflection point is \((\frac{2}{3},\frac{961}{9})\)