Question 150581
f(x)=(x+1)/(3x-1)
If f(c)=c, then
{{{(c+1)/(3c-1)=c}}}
{{{(c+1)=c(3c-1)}}}
{{{c+1=3c^2-c}}}
{{{3c^2-2c-1=0}}}
Using the quadratic formula to solve for c,
{{{c = (-(-2) +- sqrt( (-2)^2-4*3*(-1) ))/(2*3) }}}
{{{c = (2 +- sqrt( 4+12))/(6) }}}
{{{c = (2 +- sqrt( 16))/(6) }}}
{{{c = (2 +- 4)/(6) }}}
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{{{c[1] = (2 + 4)/(6) }}}
{{{c[1] = (6)/(6) }}}
{{{c[1] = 1 }}}
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{{{c[2] = (2 - 4)/(6) }}}
{{{c[2] = (-2)/(6) }}}
{{{c[2] = -(1/3) }}}