Question

let f(x)=x^3 - 7x^2 + 5x - 1. 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 $(x^n)'=nx^{n - 1}$, if $f(x)=x^{3}-7x^{2}+5x - 1$, then $f'(x)=3x^{2}-14x + 5$.

Step2: Find the second - derivative

Differentiate $f'(x)$ again. $f''(x)=(3x^{2}-14x + 5)'=6x-14$.

Step3: Find the inflection points

Set $f''(x) = 0$. So, $6x-14 = 0$. Solving for $x$ gives $x=\frac{14}{6}=\frac{7}{3}$.

Step4: Determine concavity

Test intervals based on the inflection - point $x = \frac{7}{3}$.
Choose a test point in the interval $(-\infty,\frac{7}{3})$, say $x = 0$. Then $f''(0)=6\times0 - 14=-14<0$, so $f(x)$ is concave down on $(-\infty,\frac{7}{3})$.
Choose a test point in the interval $(\frac{7}{3},\infty)$, say $x = 3$. Then $f''(3)=6\times3 - 14 = 4>0$, so $f(x)$ is concave up on $(\frac{7}{3},\infty)$.

Answer:

1. $(\frac{7}{3},\infty)$
2. $(-\infty,\frac{7}{3})$
3. $\frac{7}{3}$