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𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟒.
𝑎) 𝑆𝑒𝑗𝑎 𝑚 =
4
9
𝑘
1
3 𝑎 𝑖𝑛𝑐𝑙𝑖𝑛𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑡𝑒 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑦 = √
𝑒𝑥 + 𝑒−𝑥
𝑒𝑥 − 𝑒−𝑥
3
𝑛𝑜
𝑝𝑜𝑛𝑡𝑜 𝑥 = ln 2 . 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑘.
𝑦 = √
𝑒𝑥 + 𝑒−𝑥
𝑒𝑥 − 𝑒−𝑥
3
= √
𝑒𝑥 + 𝑒−𝑥
𝑒𝑥 − 𝑒−𝑥
.
𝑒𝑥
𝑒𝑥
3
= √
𝑒2𝑥 + 1
𝑒2𝑥 − 1
3
; 𝐷(𝑦) = {𝑥 ∈ ℝ | 𝑥 ≠ 0}
𝑃𝑎𝑟𝑎 𝑥 = ln 2 𝑡𝑒𝑚𝑜𝑠
𝑦 = √
𝑒2 ln 2 + 1
𝑒2 ln 2 − 1
3
= √
𝑒 ln 4 +1
𝑒 ln 4 −1
3
= √
4+ 1
4− 1
3
= √
5
3
3
= (
5
3
)
1
3
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒
ln|𝑦| = ln |√
𝑒2𝑥 +1
𝑒2𝑥 −1
3
| = ln |
𝑒2𝑥 + 1
𝑒2𝑥 − 1
|
1
3
=
1
3
[ln|𝑒2𝑥 + 1|− ln|𝑒2𝑥 − 1|]
𝑃𝑜𝑟 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎çã𝑜 𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑖𝑐𝑎 𝑡𝑒𝑚𝑜𝑠:
𝑦′
𝑦
=
1
3
[
2𝑒2𝑥
𝑒2𝑥 +1
−
2𝑒2𝑥
𝑒2𝑥 −1
] ⟹ 𝑦′ =
𝑦
3
[
2𝑒2𝑥
𝑒2𝑥 + 1
−
2𝑒2𝑥
𝑒2𝑥 − 1
]
𝑦′(ln2) =
1
3
. (
5
3
)
1
3
[
2.𝑒 ln 4
𝑒 ln 4 +1
−
2. 𝑒 ln 4
𝑒 ln 4 − 1
]
=
1
3
. (
5
3
)
1
3
[
2 × 4
4 + 1
−
2× 4
4− 1
]
=
1
3
. (
5
3
)
1
3
[
8
5
−
8
3
] =
1
3
.
−16
15
. (
5
3
)
1
3
= −
16
45
. (
5
3
)
1
3
𝐶𝑜𝑚𝑜 𝑚 = 𝑦′(ln2) é 𝑜 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒, 𝑒𝑛𝑡ã𝑜
4
9
𝑘
1
3 = −
16
45
. (
5
3
)
1
3
𝑘
1
3 = −
4
5
. (
5
3
)
1
3
𝑘 = −
64
75
𝑏) 𝑈𝑚 𝑐𝑖𝑙𝑖𝑛𝑑𝑟𝑜 𝑟𝑒𝑡𝑜 é 𝑔𝑒𝑟𝑎𝑑𝑜 𝑝𝑒𝑙𝑎 𝑟𝑜𝑡𝑎çã𝑜 𝑑𝑒 𝑢𝑚 𝑟𝑒𝑡â𝑛𝑔𝑢𝑙𝑜 𝑐𝑜𝑚 𝑝𝑒𝑟í𝑚𝑒𝑡𝑟𝑜 𝑝 𝑒𝑚
𝑡𝑜𝑟𝑛𝑜 𝑑𝑒 𝑢𝑚 𝑑𝑒 𝑠𝑒𝑢𝑠 𝑙𝑎𝑑𝑜𝑠. 𝑄𝑢𝑒 𝑑𝑖𝑚𝑒𝑛𝑠õ𝑒𝑠 𝑑𝑒𝑣𝑒𝑚 𝑡𝑒𝑟 𝑜 𝑟𝑒𝑡â𝑛𝑔𝑢𝑙𝑜 𝑝𝑎𝑟𝑎 𝑔𝑒𝑟𝑎𝑟 𝑜
𝑐𝑖𝑙𝑖𝑛𝑑𝑟𝑜 𝑑𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑚á𝑥𝑖𝑚𝑜?