Question

let (f(t)=\frac{-10}{(t^{2}-4t + 9)}).
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), separate by commas if needed. write none if there is no answer.)
d. find all relative minimum points. (enter ordered pair(s), separate by commas if needed. write none if there is no answer.)

Explanation:

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

First, rewrite \(f(t)=- 10(t^{2}-4t + 9)^{-1}\). Using the chain - rule \((u^{-n})^\prime=-nu^{-n - 1}u^\prime\), where \(u = t^{2}-4t + 9\) and \(n = 1\). The derivative of \(u=t^{2}-4t + 9\) is \(u^\prime=2t-4\). So \(f^\prime(t)=-10\times(-1)\times(t^{2}-4t + 9)^{-2}\times(2t - 4)=\frac{10(2t - 4)}{(t^{2}-4t + 9)^{2}}=\frac{20(t - 2)}{(t^{2}-4t + 9)^{2}}\).

Step2: Find where \(f^\prime(t)>0\) (increasing intervals)

Set \(f^\prime(t)>0\). Since \((t^{2}-4t + 9)^{2}>0\) for all real \(t\) (because \(t^{2}-4t + 9=(t - 2)^{2}+5>0\)), we only need to consider the sign of the numerator. Solve \(20(t - 2)>0\), which gives \(t>2\). So the function is increasing on the interval \((2,\infty)\).

Step3: Find where \(f^\prime(t)<0\) (decreasing intervals)

Set \(f^\prime(t)<0\). Again, since \((t^{2}-4t + 9)^{2}>0\) for all real \(t\), consider the sign of the numerator. Solve \(20(t - 2)<0\), which gives \(t<2\). So the function is decreasing on the interval \((-\infty,2)\).

Step4: Find critical points

Set \(f^\prime(t) = 0\). Since \((t^{2}-4t + 9)^{2}
eq0\) for all real \(t\), set \(20(t - 2)=0\), so \(t = 2\).

Step5: Determine relative extrema

Since \(f(t)\) changes from decreasing (\(f^\prime(t)<0\) for \(t<2\)) to increasing (\(f^\prime(t)>0\) for \(t>2\)) at \(t = 2\), there is a relative minimum at \(t = 2\). Substitute \(t = 2\) into \(f(t)\): \(f(2)=\frac{-10}{2^{2}-4\times2 + 9}=\frac{-10}{4-8 + 9}=\frac{-10}{5}=-2\). The relative - minimum point is \((2,-2)\). There are no relative maximum points.

Answer:

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