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

Given $f(x)=x^{3}-5x^{2}+3x + 2$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $f'(x)=3x^{2}-10x + 3$.

Step2: Find the second - derivative

Differentiate $f'(x)$ with respect to $x$. $f''(x)=\frac{d}{dx}(3x^{2}-10x + 3)=6x-10$.

Step3: Find intervals of concavity

1. For concave up:

Set $f''(x)>0$. Then $6x - 10>0$. Solving for $x$, we get $6x>10$, so $x>\frac{5}{3}$. The interval where $f(x)$ is concave up is $(\frac{5}{3},\infty)$.

1. For concave down:

Set $f''(x)<0$. Then $6x - 10<0$. Solving for $x$, we get $6x<10$, so $x<\frac{5}{3}$. The interval where $f(x)$ is concave down is $(-\infty,\frac{5}{3})$.

Step4: Find inflection points

Set $f''(x) = 0$. Then $6x-10 = 0$. Solving for $x$, we get $6x=10$, so $x=\frac{5}{3}$.

Answer:

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