Question 1133690

 Given that
 
{{{f(x)=ax+b}}},
 
{{{f(1)=1}}} and

{{{ f ^-1(5)=2}}},


 find value of {{{a }}}and of{{{ b}}}. 


if {{{f(1)=1}}}=>{{{f(x)=1}}} when {{{x=1}}}

then {{{1=a*1+b}}}=>{{{1=a+b}}}=>solve for {{{a}}}

{{{1-b=a}}}....eq.1


now first find {{{ f ^-1(x)}}}

{{{f(x)=y}}}
{{{y=ax+b}}}.....swap {{{x}}} and {{{y}}}
{{{x=ay+b}}}......solve for {{{y}}}
{{{x-b=ay}}}
{{{y=(x-b)/a}}}.........eq.2


=>{{{ f ^-1(x)=(x-b)/a}}}

if {{{ f ^-1(5)=2}}}, we have

{{{ 2=(5-b)/a}}}

{{{ a=(5-b)/2}}}........eq.3


from eq.1 and eq.3 we have:


{{{1-b=(5-b)/2}}}......solve for {{{b}}}

{{{2(1-b)=5-b}}}

{{{2-2b=5-b}}}

{{{2-5=2b-b}}}

{{{b=-3}}}


now find {{{a}}}

{{{a=1-(-3)}}}....eq.1

{{{a=1+3}}}

{{{a=4}}}


so, your equation is: {{{f(x)=4x-3}}}


now, go to eq.2, plug in {{{a}}} and {{{b}}}

{{{ f ^-1(x)=(x-b)/a}}}

{{{ f ^-1(x)=(x-(-3))/4}}}

{{{ f ^-1(x)=(x+3)/4}}}

{{{ f ^-1(x)=(1/4)x+3/4}}}


and evaluate
 
{{{f^-1(7)=(1/4)7+3/4=7/4+3/4=10/4=5/2}}}


 and {{{f ^-1}}}({{{-5}}}{{{ 1/2}}})=. ({{{-11/2}}})

{{{f^-1(-11/2)=(1/4)(-11/2)+3/4=-11/8+3/4=-11/8+6/8=-5/8}}}