let (f(x)=-x^{3}+15x^{2}-2x + 9). a. find the inflection point. enter your answer as an ordered pair. b…

let (f(x)=-x^{3}+15x^{2}-2x + 9). a. find the inflection point. enter your answer as an ordered pair. b. identify interval(s) where the graph of (f(x)) is concave up. write none if there is no answer. c. identify interval(s) where the graph of (f(x)) is concave down. write none if there is no answer.

let (f(x)=-x^{3}+15x^{2}-2x + 9). a. find the inflection point. enter your answer as an ordered pair. b. identify interval(s) where the graph of (f(x)) is concave up. write none if there is no answer. c. identify interval(s) where the graph of (f(x)) is concave down. write none if there is no answer.

Answer

Explanation:

Step1: Find the first - derivative

Given $f(x)=-x^{3}+15x^{2}-2x + 9$. Using the power rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=-3x^{2}+30x - 2$.

Step2: Find the second - derivative

Differentiate $f'(x)$ with respect to $x$. So $f''(x)=-6x + 30$.

Step3: Find the inflection point

Set $f''(x) = 0$. Then $-6x+30 = 0$. Solving for $x$ gives $x = 5$. Substitute $x = 5$ into $f(x)$: $f(5)=-5^{3}+15\times5^{2}-2\times5 + 9=-125 + 375-10 + 9=249$. The inflection point is $(5,249)$.

Step4: Determine concavity intervals

Test intervals for $f''(x)$. Consider the intervals $(-\infty,5)$ and $(5,\infty)$. For $x<5$, let's take $x = 0$. Then $f''(0)=-6\times0 + 30=30>0$, so the function is concave up on the interval $(-\infty,5)$. For $x>5$, let's take $x = 6$. Then $f''(6)=-6\times6 + 30=-6<0$, so the function is concave down on the interval $(5,\infty)$.

Answer:

a. $(5,249)$ b. $(-\infty,5)$ c. $(5,\infty)$