Quiz 6 Key
1. Determine the critical numbers for each of the following functions:
a. 𝑓(𝑥)=2𝑥3−5𝑥2−7𝑥+6
Note that the domain is all real numbers.
𝑓′(𝑥)=6𝑥2−10𝑥−7
𝑓′(𝑥) is never undefined.
Solving 𝑓′(𝑥)=0⇒6𝑥2−10𝑥−7=0⇒𝑥=10±√268
12 =10±2√67
12 =5±√67
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So the critical numbers of 𝑓(𝑥) are 𝑥=5±√67
6
b. 𝑠(𝑡)=√𝑡2−6𝑡+8
3+4
Note that the domain is all real numbers. [We have an odd root here, so no worries about roots of negative numbers.]
𝑠′(𝑡)=13(𝑡2−6𝑡+8)−23∙(2𝑡−6)=2𝑡−6
13(𝑡2−6𝑡+8)23=2𝑡−6
3√(𝑡2−6𝑡+8)2
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Note that 𝑠′(𝑡) will be undefined if 𝑡2−6𝑡+8=0⇒(𝑡−2)(𝑡−4)=0⇒𝑡=2,𝑡=4
Solving 𝑠′(𝑡)=0⇒2𝑡−6=0⇒𝑡=3
So the critical numbers of 𝑠(𝑡) are 𝑡=2,3,4
c. ℎ(𝑝)=8cos(3𝑝)+12𝑝, for 0≤𝑝≤2𝜋
Note that the domain is all real numbers.
ℎ′(𝑝)=8(−sin(3𝑝)∙3)+12=−24sin(3𝑝)+12
ℎ′(𝑝) is never undefined.
Solving ℎ′(𝑝)= 0⇒ =−24sin(3𝑝)+12=0⇒sin(3𝑝)=12⇒ 3𝑝= 𝜋6,5𝜋
6,13𝜋
6,17𝜋
6,…
So 𝑝=𝜋
18,5𝜋
18,13𝜋
18,17𝜋
18,…
Considering the stated restricted domain of [0,2𝜋], then the critical numbers of ℎ(𝑝) are 𝑝=𝜋
18,5𝜋
18,13𝜋
18,17𝜋
18,25𝜋
18,29𝜋
18
2. Determine the absolute maximum and minimum values for each function over the given interval. Note that
these are the same functions used in exercise 1. You may use your results from exercise 1.
a. 𝑓(𝑥)=2𝑥3−5𝑥2−7𝑥+6 on [0,4]
Evaluating 𝑓(𝑥) at 𝑥=0,5+√67
6,4
𝑓(0)=6
𝑓(5+√67
6)≈−12.30405
𝑓(4)=26
So, on the given interval, 𝑓(𝑥) has an absolute maximum of 26 when 𝑥=4 and an absolute minimum of approximately
-12.30405 when 𝑥=5+√67
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