Question

let (f(t)=\frac{-6}{(t^{2}-4t + 5)}).
a. identify interval(s) where the graph of (f(t)) is increasing. (separate intervals with commas if needed.)
b. identify interval(s) where the graph of (f(t)) is decreasing. (separate intervals with commas if needed.)
c. find all relative maximum points. (enter ordered pair(s), separated by commas if needed. write none if there is no answer.)
d. find all relative minimum points. (enter ordered pair(s), separated by commas if needed. write none if there is no answer.)

Explanation:

Step1: Find the derivative of $f(t)$

Use the quotient - rule. If $f(t)=\frac{u}{v}$ where $u = - 6$ and $v=t^{2}-4t + 5$, then $u'=0$ and $v'=2t - 4$. The quotient - rule states that $f'(t)=\frac{u'v - uv'}{v^{2}}$. So $f'(t)=\frac{0\times(t^{2}-4t + 5)-(-6)\times(2t - 4)}{(t^{2}-4t + 5)^{2}}=\frac{12t-24}{(t^{2}-4t + 5)^{2}}$.

Step2: Find the critical points

Set $f'(t) = 0$. Since the denominator $(t^{2}-4t + 5)^{2}>0$ for all real $t$ (because $t^{2}-4t + 5=(t - 2)^{2}+1>0$ for all $t\in R$), we set the numerator equal to 0. So $12t-24 = 0$, which gives $t = 2$.

Step3: Test the intervals

Choose test points in the intervals $(-\infty,2)$ and $(2,\infty)$. Let's choose $t = 1$ for the interval $(-\infty,2)$ and $t=3$ for the interval $(2,\infty)$.
For $t = 1$, $f'(1)=\frac{12\times1-24}{(1^{2}-4\times1 + 5)^{2}}=\frac{12 - 24}{(1 - 4 + 5)^{2}}=\frac{-12}{4}=-3<0$.
For $t = 3$, $f'(3)=\frac{12\times3-24}{(3^{2}-4\times3 + 5)^{2}}=\frac{36-24}{(9 - 12 + 5)^{2}}=\frac{12}{4}=3>0$.

a. The function $f(t)$ is increasing on the interval $(2,\infty)$.
b. The function $f(t)$ is decreasing on the interval $(-\infty,2)$.
c. Since the function changes from decreasing to increasing at $t = 2$, the relative minimum point is $(2,f(2))$. $f(2)=\frac{-6}{2^{2}-4\times2 + 5}=\frac{-6}{4 - 8 + 5}=-6$. So the relative minimum point is $(2,-6)$, and there are no relative maximum points. So the answer for c is None.
d. The relative minimum point is $(2,-6)$.

Answer:

a. $(2,\infty)$
b. $(-\infty,2)$
c. None
d. $(2,-6)$