Exercicios Calculo (78)

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90 
 
 
𝐶𝑜𝑚 𝑥 =
𝜋
4
 𝑒 ∆𝑥 = 𝑑𝑥 = −
𝜋
180
, 𝑜𝑏𝑡𝑒𝑚𝑜𝑠: 
 
𝑓 (
𝜋
4
−
𝜋
180
) − 𝑓 (
𝜋
4
) ≅ 𝑓 ′ (
𝜋
4
) . (−
𝜋
180
) 
 
𝑓 (
𝜋
4
−
𝜋
180
) − 1 ≅ 2.(−
𝜋
180
) 
 
𝑓 (
𝜋
4
−
𝜋
180
) ≅ 1−
𝜋
90
 
 
𝐶𝑜𝑚𝑜 𝑓(
𝜋
4
−
𝜋
180
) = tg 44° , 𝑒𝑛𝑡ã𝑜 tg 44° ≅ 1 −
𝜋
90
. 
 
𝑸𝒖𝒆𝒔𝒕ã𝒐 𝟑. 
 
𝑎) 𝑆𝑒𝑗𝑎 𝑓 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖á𝑣𝑒𝑙 𝑒𝑚 (0,1) 𝑒 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 [0,1] 𝑐𝑜𝑚 𝑓(0) = 0 𝑒 𝑓(1) = 1. 
𝑀𝑜𝑠𝑡𝑟𝑒 𝑞𝑢𝑒 𝑒𝑥𝑖𝑠𝑡𝑒 𝑐 ∈ (0,1) 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓(𝑐) =
1
2
.𝑈𝑠𝑒 𝑖𝑠𝑡𝑜 𝑝𝑎𝑟𝑎 𝑚𝑜𝑠𝑡𝑟𝑎𝑟 𝑞𝑢𝑒 𝑒𝑥𝑖𝑠𝑡𝑒𝑚 
𝑥 𝑒 𝑦 𝑒𝑚 (0,1) 𝑡𝑎𝑖𝑠 𝑞𝑢𝑒 
1
𝑓′(𝑥)
+
1
𝑓′(𝑦)
= 2. 
𝐶𝑜𝑚𝑜 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑓𝑒𝑐ℎ𝑎𝑑𝑜 [0,1] 𝑒 
1
2
 é 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑒𝑛𝑡𝑟𝑒 𝑓(0) 𝑒 𝑓(1), 
𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖á𝑟𝑖𝑜,𝑒𝑥𝑖𝑠𝑡𝑒 𝑎𝑙𝑔𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑐 ∈ (0,1) 𝑡𝑎𝑙 𝑞𝑢𝑒 
𝑓(𝑐) =
1
2
. 
 
𝑓 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑞𝑢𝑒 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑧 𝑎𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 ℎ𝑖𝑝ó𝑡𝑒𝑠𝑒𝑠: 
 
1. 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 [0,𝑐] 𝑒 𝑒𝑚 [𝑐, 1]; 
2. 𝑓 é 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑒𝑚 (0,𝑐) 𝑒 𝑒𝑚 (𝑐, 1); 
 
𝐸𝑛𝑡ã𝑜, 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝑀é𝑑𝑖𝑜 𝑒𝑥𝑖𝑠𝑡𝑒𝑚 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑥 𝑒 𝑦, 𝑐𝑜𝑚 𝑥 ∈ (0,𝑐) 𝑒 
𝑦 ∈ (𝑐, 1) 𝑡𝑎𝑖𝑠 𝑞𝑢𝑒: 
 
𝑓 ′(𝑥) =
𝑓(𝑐) − 𝑓(0)
𝑐 − 0
=
1
2 − 0
𝑐
=
1
2𝑐
. 
 
𝑓 ′(𝑦) =
𝑓(1) − 𝑓(𝑐)
1 − 𝑐
=
1−
1
2
1 − 𝑐
=
1
2(1 − 𝑐)
. 
 
1
𝑓 ′(𝑥)
+
1
𝑓 ′(𝑦)
=
1
1
2𝑐
+
1
1
2(1− 𝑐)
= 2𝑐 + 2(1 − 𝑐) = 2𝑐 + 2− 2𝑐 = 2. 
 
𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜,
1
𝑓 ′(𝑥)
+
1
𝑓 ′(𝑦)
= 2.