Dada a função expoencial f(x)=4^x,determine:
Anexos:
![](https://pt-static.z-dn.net/files/dc4/fe0c93e25b1174ec36ad2a2529994ffa.jpg)
Soluções para a tarefa
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a) ![f(0) f(0)](https://tex.z-dn.net/?f=f%280%29)
![f(x) = 4^x \\ f(0)=4^0\\f(0)=1 f(x) = 4^x \\ f(0)=4^0\\f(0)=1](https://tex.z-dn.net/?f=f%28x%29+%3D+4%5Ex+%5C%5C+f%280%29%3D4%5E0%5C%5Cf%280%29%3D1)
b)![f(3) f(3)](https://tex.z-dn.net/?f=f%283%29)
![f(x) = 4^x \\ f(3)=4^3\\f(3)=4*4*4=64 f(x) = 4^x \\ f(3)=4^3\\f(3)=4*4*4=64](https://tex.z-dn.net/?f=f%28x%29+%3D+4%5Ex+%5C%5C+f%283%29%3D4%5E3%5C%5Cf%283%29%3D4%2A4%2A4%3D64)
c)![f(-1) f(-1)](https://tex.z-dn.net/?f=f%28-1%29)
![f(x) = 4^x \\ f(-1)=4^{-1}\\f(-1)= \frac{1}{4} = 0,25 f(x) = 4^x \\ f(-1)=4^{-1}\\f(-1)= \frac{1}{4} = 0,25](https://tex.z-dn.net/?f=f%28x%29+%3D+4%5Ex+%5C%5C+f%28-1%29%3D4%5E%7B-1%7D%5C%5Cf%28-1%29%3D+%5Cfrac%7B1%7D%7B4%7D+%3D+0%2C25)
d)![f( \frac{3}{2} ) f( \frac{3}{2} )](https://tex.z-dn.net/?f=f%28+%5Cfrac%7B3%7D%7B2%7D+%29)
![f(x) = 4^x \\ f( \frac{3}{2} )=4 ^{\frac{3}{2}} = \sqrt[2]{4^3} = \sqrt[2]{64} = 8 f(x) = 4^x \\ f( \frac{3}{2} )=4 ^{\frac{3}{2}} = \sqrt[2]{4^3} = \sqrt[2]{64} = 8](https://tex.z-dn.net/?f=f%28x%29+%3D+4%5Ex+%5C%5C+f%28+%5Cfrac%7B3%7D%7B2%7D+%29%3D4+%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D+%3D++%5Csqrt%5B2%5D%7B4%5E3%7D+%3D++%5Csqrt%5B2%5D%7B64%7D+%3D+8)
e)![f( -\frac{1}{2} ) f( -\frac{1}{2} )](https://tex.z-dn.net/?f=f%28+-%5Cfrac%7B1%7D%7B2%7D+%29)
![f(x) = 4^x \\ f(- \frac{1}{2} )=4 ^{-\frac{1}{2}} = (\frac{1}{4} )^{ \frac{1}{2}}= \frac{ \sqrt{1}}{ \sqrt{4} } = \frac{1}{2} = 0,5 f(x) = 4^x \\ f(- \frac{1}{2} )=4 ^{-\frac{1}{2}} = (\frac{1}{4} )^{ \frac{1}{2}}= \frac{ \sqrt{1}}{ \sqrt{4} } = \frac{1}{2} = 0,5](https://tex.z-dn.net/?f=f%28x%29+%3D+4%5Ex+%5C%5C+f%28-+%5Cfrac%7B1%7D%7B2%7D+%29%3D4+%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D+%3D++%28%5Cfrac%7B1%7D%7B4%7D+%29%5E%7B+%5Cfrac%7B1%7D%7B2%7D%7D%3D+%5Cfrac%7B+%5Csqrt%7B1%7D%7D%7B+%5Csqrt%7B4%7D+%7D+%3D++%5Cfrac%7B1%7D%7B2%7D+%3D+0%2C5)
f) m tal que![f(m)=1 f(m)=1](https://tex.z-dn.net/?f=f%28m%29%3D1)
![f(x)=4^x \\ f(m)=4^m f(x)=4^x \\ f(m)=4^m](https://tex.z-dn.net/?f=f%28x%29%3D4%5Ex+%5C%5C+f%28m%29%3D4%5Em)
Como ele quer que
, substituímos na equação.
![f(m)=4^m \\ f(m)=1 \\\\ 4^m=1 f(m)=4^m \\ f(m)=1 \\\\ 4^m=1](https://tex.z-dn.net/?f=f%28m%29%3D4%5Em+%5C%5C+f%28m%29%3D1+%5C%5C%5C%5C+4%5Em%3D1)
Como precisamos igualar as bases, e
, teremos:
![4^m = 1 \\ 4^m = 4^0 \\ m = 0 4^m = 1 \\ 4^m = 4^0 \\ m = 0](https://tex.z-dn.net/?f=4%5Em+%3D+1+%5C%5C+4%5Em+%3D+4%5E0+%5C%5C+m+%3D+0)
b)
c)
d)
e)
f) m tal que
Como ele quer que
Como precisamos igualar as bases, e
viicxxxxo:
Muito obrigada!
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