Question

sendo f(x)=a^x, com a>0 sabendo que f(1) +f(-1)= -5/2, calcule f(2)

Answer 1

$f(x)=a^x$

$f(1)=a$

$f(-1)=\dfrac{1}{a}$

$f(1)+f(-1)=-\dfrac{5}{2}$

$a+\dfrac{1}{a}=-\dfrac{5}{2}$

$\dfrac{a^2+1}{a}=\dfrac{-5}{2}$

$2a^2+2=-5a$

$2a^2+5a+2=0$

$\Delta=5^2-4\cdot2\cdot2=25-16=9$

$a=\dfrac{-5\pm\sqrt{9}}{2\cdot2}=\dfrac{-5\pm3}{4}$

$a'=\dfrac{-5+3}{4}=\dfrac{-2}{4}=-\dfrac{1}{2}$

$a"=\dfrac{-5-3}{4}=\dfrac{-8}{4}=-2$

$f(x)=(-2)^x$

$f(2)=(-2)^2$

$f(2)=4$

Answer 2

$\boxed{f(x)=a^x}\\ \\ f(1)+f(-1)=-\frac{5}{2}\\ \\ a^1+a^{-1}=-\frac{5}{2}\\ \\a+\frac{1}{a}=-\frac{5}{2}\\ \\2a^2+2=-5a\\ \\2a^2+5a+2=0\\ \\\Delta=5^2-4.2.2=25-16=9\\\\ a=\frac{-5\pm3}{4}\\\\ a_1=-2\\\\ a_2=-\frac{1}{2}$

Se a= -2

$f(2)=(-2)^2=4$

Se a = -1/2

$f(2)=(-\frac{1}{2})^2=\frac{1}{4}$