Question

let (f(x)=x^{3}+3x^{2}-9x + 9). (a) use the definition of a derivative or the derivative rules to find (f(x)=) (b) use the definition of a derivative or the derivative rules to find (f(x)=) (c) on what interval(s) is (f) increasing? (x:) (d) on what interval(s) is (f) decreasing? (x:) (e) on what interval(s) is (f) concave downward? (x:) (f) on what interval(s) is (f) concave upward? (x:) question help: video

Explanation:

Step1: Find the first - derivative using power rule

The power rule states that if $y = x^n$, then $y'=nx^{n - 1}$. For $f(x)=x^{3}+3x^{2}-9x + 9$, we have $f'(x)=3x^{2}+6x - 9$.

Step2: Find the second - derivative using power rule

Differentiate $f'(x)=3x^{2}+6x - 9$ with respect to $x$. Using the power rule, $f''(x)=6x + 6$.

Step3: Find critical points of $f(x)$

Set $f'(x)=0$. So, $3x^{2}+6x - 9 = 0$. Divide through by 3: $x^{2}+2x - 3=0$. Factor: $(x + 3)(x - 1)=0$. The critical points are $x=-3$ and $x = 1$.

Step4: Determine intervals of increase and decrease

Test intervals $(-\infty,-3)$, $(-3,1)$ and $(1,\infty)$ using $f'(x)$.
For $x<-3$, let $x=-4$, then $f'(-4)=3\times(-4)^{2}+6\times(-4)-9=48-24 - 9=15>0$.
For $-3For $x>1$, let $x = 2$, then $f'(2)=3\times2^{2}+6\times2-9=12 + 12-9=15>0$.
So, $f(x)$ is increasing on $(-\infty,-3)\cup(1,\infty)$ and decreasing on $(-3,1)$.

Step5: Find inflection point

Set $f''(x)=0$. So, $6x + 6=0$, which gives $x=-1$.

Step6: Determine intervals of concavity

Test intervals $(-\infty,-1)$ and $(-1,\infty)$ using $f''(x)$.
For $x<-1$, let $x=-2$, then $f''(-2)=6\times(-2)+6=-6<0$.
For $x>-1$, let $x = 0$, then $f''(0)=6\times0+6=6>0$.
So, $f(x)$ is concave downward on $(-\infty,-1)$ and concave upward on $(-1,\infty)$.

Answer:

(a) $f'(x)=3x^{2}+6x - 9$
(b) $f''(x)=6x + 6$
(c) $x\in(-\infty,-3)\cup(1,\infty)$
(d) $x\in(-3,1)$
(e) $x\in(-\infty,-1)$
(f) $x\in(-1,\infty)$