Question 1090249

1.)
{{{f(x)=2(x-8)^2+5}}} what is {{{f^-1(7)}}}

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

{{{f(x)=y=2(x-8)^2+5}}}

{{{y=2(x-8)^2+5}}}...........swap {{{x}}} and {{{y}}}

{{{x=2(y-8)^2+5}}}

{{{x-5=2(y-8)^2}}}

{{{(x-5)/2=(y-8)^2}}}

{{{sqrt((x-5)/2)=y-8}}}

{{{y}}}=± {{{sqrt((x-5)/2)+8}}}


{{{f^-1(x)}}}=± {{{sqrt((x-5)/2)+8}}}

now find {{{f^-1(7)}}}

{{{f^-1(7)}}}=± {{{sqrt((7-5)/2)+8}}}

{{{f^-1(7)}}}=± {{{9}}}