Question 941055
{{{f^(-1)(6)=3}}} so {{{f(3)=6}}}
{{{f^(-1)(29)=-2}}} so {{{f(-2)=29}}}
{{{f(x)=px+q}}}
1.{{{f(3)=3p+q=6}}}
2.{{{f(-2)=-2p+q=29}}}
Subtract eq. 1 from eq. 2,
{{{-2p+q-3p-q=29-6}}}
{{{-5p=23}}}
{{{p=-23/5}}}
Then use either equation to solve for {{{q}}}
{{{3(-23/5)+q=6}}}
{{{-69/5+q=30/5}}}
{{{q=30/5+69/5}}}
{{{q=99/5}}}
So then,
{{{f(x)=(-23/5)x+99/5}}}
{{{-(23/5)x+99/5=27}}}
{{{-(23/5)x=135/5-99/5}}}
{{{-(23/5)x=36/5}}}
{{{x=-36/23}}}
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The answer of {{{6}}} doesn't make sense.
Since the solution is linear and {{{27}}} is between {{{6}}} and {{{29}}}, the solution needs to be between {{{3}}} and {{{-2}}}, closer to {{{-2}}} since {{{27}}} is closer to {{{29}}}. My solution is about {{{-1.5}}}, close compared to {{{-2}}}. {{{6}}} is not even in the possible range of solutions.