Question

let f(x)=x^3 - 5x^2 + 3x + 2. find the open intervals on which f is concave up (down). then determine the x - coordinates of all inflection points of f. 1. f is concave up on the intervals 2. f is concave down on the intervals 3. the inflection points occur at x = notes: in the first two, your answer should either be a single interval, such as (0,1), a comma separated list of intervals, such as (-inf, 2), (3,4), or the word
one\. in the last one, your answer should be a comma separated list of x values or the word
one\.

Explanation:

Step1: Find the first - derivative

Using the power rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=3x^{2}-10x + 3$.

Step2: Find the second - derivative

Differentiate $f'(x)$ again. $f''(x)=6x-10$.

Step3: Find the inflection points

Set $f''(x) = 0$. Then $6x-10 = 0$, which gives $x=\frac{10}{6}=\frac{5}{3}$.

Step4: Determine concavity

Test intervals based on the inflection - point $x = \frac{5}{3}$.
For $x<\frac{5}{3}$, let's take $x = 1$. Then $f''(1)=6\times1 - 10=-4<0$, so $f(x)$ is concave down on the interval $(-\infty,\frac{5}{3})$.
For $x>\frac{5}{3}$, let's take $x = 2$. Then $f''(2)=6\times2 - 10 = 2>0$, so $f(x)$ is concave up on the interval $(\frac{5}{3},\infty)$.

Answer:

1. $(\frac{5}{3},\infty)$
2. $(-\infty,\frac{5}{3})$
3. $\frac{5}{3}$