Exercicios Calculo (196)

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213 
 
𝐴 𝑝𝑟𝑖𝑚𝑖𝑡𝑖𝑣𝑎 𝑚𝑎𝑖𝑠 𝑔𝑒𝑟𝑎𝑙 𝑑𝑒 𝑓 ′é 𝑑𝑎𝑑𝑎 𝑝𝑜𝑟 𝑓(𝑥) = −
1
2
cos(2𝑥) + tg 𝑥 + 𝐶, 𝑜𝑛𝑑𝑒 𝐶 
é 𝑢𝑚𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒. 
 
𝑓 (
𝜋
6
) =
√3
3
⟹ −
1
2
cos(
𝜋
3
) + tg (
𝜋
6
) + 𝐶 =
√3
3
⇒ −
1
4
+
√3
3
+ 𝐶 =
√3
3
∴ 𝐶 =
1
4
. 
 
𝐿𝑜𝑔𝑜, 𝑓(𝑥) = −
1
2
cos(2𝑥)+ tg 𝑥 +
1
4
. 
 
𝑏) 𝑆𝑒𝑗𝑎𝑚 𝑥,𝑦 𝑟𝑒𝑎𝑖𝑠 𝑛ã𝑜 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑜𝑠,𝑡𝑎𝑖𝑠 𝑞𝑢𝑒 𝑥 + 𝑦 = 1,𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑒 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑚í𝑛𝑖𝑚𝑜 𝑑𝑒 
𝑥
3
2 + 𝑦
3
2 . 
𝑦 = 1− 𝑥 
 𝑓(𝑥) = 𝑥
3
2 + (1 − 𝑥)
3
2 𝐷(𝑓) = {𝑥 ∈ ℝ | 0 ≤ 𝑥 ≤ 1} 
 
𝑓 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑓𝑒𝑐ℎ𝑎𝑑𝑜 [0,1] 𝑒,𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 
𝐸𝑥𝑡𝑟𝑒𝑚𝑜, 𝑓 𝑎𝑠𝑠𝑢𝑚𝑒 𝑢𝑚 𝑣𝑎𝑙𝑜𝑟 𝑚á𝑥𝑖𝑚𝑜 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜 𝑓(𝑐) 𝑒 𝑢𝑚 𝑣𝑎𝑙𝑜𝑟 𝑚í𝑛𝑖𝑚𝑜 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜 
𝑓(𝑑) 𝑒𝑚 𝑐𝑒𝑟𝑡𝑜𝑠 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑐 𝑒 𝑑 𝑒𝑚 [0,1]. 
 
𝑃𝑒𝑙𝑜 𝑀é𝑡𝑜𝑑𝑜 𝑑𝑜 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝐹𝑒𝑐ℎ𝑎𝑑𝑜 𝑡𝑒𝑚𝑜𝑠 
 
1. 𝑂𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑓 𝑛𝑜𝑠 𝑒𝑥𝑡𝑟𝑒𝑚𝑜𝑠 𝑑𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜: 
 
𝑓(0) = 0
3
2 +1
3
2 = 1. 
𝑓(1) = 1
3
2 +0
3
2 = 1. 
 
2. 𝑂𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑓 𝑛𝑜𝑠 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑐𝑟í𝑡𝑖𝑐𝑜𝑠 𝑑𝑒 𝑓 𝑒𝑚 (0,1): 
 
"𝑈𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑐𝑟í𝑡𝑖𝑐𝑜 𝑑𝑒 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑓 é 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑐 𝑛𝑜 𝑑𝑜𝑚í𝑛𝑖𝑜 𝑑𝑒 𝑓 𝑜𝑛𝑑𝑒 
𝑜𝑢 𝑓 ′(𝑐) = 0 𝑜𝑢 𝑜𝑛𝑑𝑒 𝑓 ′(𝑐) 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒. " 
 
𝑓 ′(𝑥) =
3
2
𝑥
1
2 −
3
2
(1 − 𝑥)
1
2 ; 𝐷(𝑓 ′) = 𝐷(𝑓) 
 
𝐶𝑜𝑚𝑜 𝑓 é 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑒𝑚 (0,1),𝑒𝑛𝑡ã𝑜 𝑠𝑒 𝑓 𝑝𝑜𝑠𝑠𝑢𝑖 𝑛ú𝑚𝑒𝑟𝑜 𝑐𝑟í𝑡𝑖𝑐𝑜 𝑐, 𝑒𝑛𝑡ã𝑜 𝑓 ′(𝑐) 
𝑒𝑥𝑖𝑠𝑡𝑒 𝑒 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜 𝑓 ′(𝑐) = 0. 
 
𝑓 ′(𝑥) = 0⟺ 𝑥
1
2 = (1− 𝑥)
1
2 ⇒ (𝑥
1
2)
2
= [(1 − 𝑥)
1
2]
2
⇒ 𝑥 = 1− 𝑥 ⇒ 2𝑥 = 1 ∴ 𝑥 =
1
2
 
 
𝑓 (
1
2
) = (
1
2
)
3
2
+ (
1
2
)
3
2
= 2 (
1
2
)
3
2
= 2.
1
√8
=
1
√2
=
√2
2
. 
 
𝐶𝑜𝑚𝑝𝑎𝑟𝑎𝑛𝑑𝑜 𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑜𝑏𝑡𝑖𝑑𝑜𝑠 𝑛𝑎𝑠 𝑒𝑡𝑎𝑝𝑎𝑠 1 𝑒 2,𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑚𝑜𝑠 𝑞𝑢𝑒 √2 2⁄ é 𝑜 
𝑣𝑎𝑙𝑜𝑟 𝑚í𝑛𝑖𝑚𝑜 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜 𝑑𝑒 𝑓 𝑒𝑚 [0,1] 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑎 𝑠𝑜𝑚𝑎 𝑥
3
2 +𝑦
3
2 é 𝑚í𝑛𝑖𝑚𝑎 𝑞𝑢𝑎𝑛𝑑𝑜 
𝑥 = 𝑦 =
1
2
.