algebra

oni asked 3 weeks ago

The function f is given by f(x)=(3^x+1)/(3^x-3^(-x) ) , for x>0
Show that f(x) > 1 for all x > 0
Solve the equation f(x) = 4

Solution:

i)Show that f(x) > 1 for all x > 0

Let us  assume  3^x =p

=> 3^{-x}=\dfrac {1}{p}

When  x=0, the value  of  p=1

Therefore we can conclude that when x > 0, p>1 always.

f\left( x\right) =\dfrac {3^{x}+1}{3^{x}-3^{-x}}

=>  f\left( x\right) =\dfrac {p+1}{p-\dfrac {1}{p}}=\dfrac {p+1}{\dfrac {p^{2}-1}{p}}

\Rightarrow f\left( x\right) =\dfrac {p\left( p+1\right) }{p^{2}-1}=\dfrac {p^{2}+p}{p^{2}-1}

Since  p>1  when x >0

we can conclude from the expression that f(x) >1

ii)Solve the equation f(x) = 4

Given f(x) = 4

=>  \begin{aligned}\Rightarrow 4=\dfrac {p^{2}+P}{p^{2}-1}\\ \Rightarrow 4\left( p^{2}-1\right) =p^{2}+p\end{aligned}

\begin{aligned}\Rightarrow 4p^{2}-4=p^{2}+p\\ \Rightarrow 3p^{2}-p-4=0\\ \Rightarrow \left( 3p-4\right) \left( p+1\right) =0\\ \Rightarrow p=-1,p=\dfrac {4}{3}\end{aligned}

Since p is always >0 , we need to  take the  solution as  p=4/3

P= 3^x we assumed initially

Therefore 3^x = 4/3

x = \ln _{3}\left( \dfrac {4}{3}\right)

 

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