Question

let f(x) be defined as f(x)=x^{3}-7x^{2}+6x + 4. use the function f to answer the following. a. find the open interval(s) on which f is concave upward. use interval notation, or enter dne if no such intervals exist. include the union symbol when entering multiple intervals. open interval(s) on which f is concave upward: b. find the open interval(s) on which f is concave downward. use interval notation, or enter dne if no such intervals exist. include the union symbol when entering multiple intervals. open interval(s) on which f is concave downward: c. find the inflection point(s). enter dne if no inflection points exist. if there is more than one point, enter the points in a comma - separated list. inflection point(s):

Explanation:

Step1: Find the first - derivative

Using the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $f(x)=x^{3}-7x^{2}+6x + 4$, we have $f'(x)=3x^{2}-14x + 6$.

Step2: Find the second - derivative

Differentiate $f'(x)$ again. $f''(x)=\frac{d}{dx}(3x^{2}-14x + 6)=6x-14$.

Step3: Find the inflection - point candidate

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

Step4: Test the intervals for concavity

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 downward 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 upward on $(\frac{7}{3},\infty)$.

Answer:

a. $(\frac{7}{3},\infty)$
b. $(-\infty,\frac{7}{3})$
c. $(\frac{7}{3},f(\frac{7}{3}))$ where $f(\frac{7}{3})=(\frac{7}{3})^{3}-7(\frac{7}{3})^{2}+6\times\frac{7}{3}+4=\frac{343}{27}-\frac{343}{9}+14 + 4=\frac{343-1029}{27}+18=\frac{-686}{27}+18=\frac{-686 + 486}{27}=-\frac{200}{27}$. So the inflection point is $(\frac{7}{3},-\frac{200}{27})$