Question 1125606
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If f(x) is one-to-one, then {{{f(a)=f(b)}}} implies that {{{a = b}}} for all a,b in the domain of f(x). Put in plain english: if two output values are the same, then the inputs must be the same for the function to be one-to-one.


Consider a counter-example such as a parabola. We can have the same output lead to two different inputs (eg: y = 4 lead to x = -2 and x = 2 for the function y = x^2). This is a reason why a parabola is not one-to-one.


{{{f(x) = 10-root(3,x-8)}}}


{{{f(a) = 10-root(3,a-8)}}}


{{{f(b) = 10-root(3,b-8)}}}


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If {{{f(a) = f(b)}}}, then...


{{{f(a) = f(b)}}}


{{{10-root(3,a-8) = 10-root(3,b-8)}}} Substitution


{{{-root(3,a-8) = -root(3,b-8)}}} Subtract 10 from both sides


{{{root(3,a-8) = root(3,b-8)}}} Multiply both sides by -1


{{{a-8 = b-8}}} Cube both sides


{{{a = b}}} Add 8 to each side


So we end up with {{{a = b}}} after assuming {{{f(a) = f(b)}}}


So this means that {{{f(a) = f(b)}}} leads to {{{a = b}}}. If we follow the steps shown above in reverse, then we'll go from {{{a = b}}} to {{{f(a) = f(b)}}}


Therefore, we have proven that f(x) is indeed one-to-one.


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Let's find the inverse. Which I'll call g(x)


g(x) = f^(-1)(x)


{{{f(x) = 10-root(3,x-8)}}}


{{{y = 10-root(3,x-8)}}} Replace f(x) with y


{{{x = 10-root(3,y-8)}}} Swap x and y. From here on out, we're solving for y.


{{{x-10 = -root(3,y-8)}}} Subtract 10 from both sides


{{{-root(3,y-8) = x-10}}} 


{{{root(3,y-8) = -x+10}}} Multiply both sides by -1


{{{y-8 = (-x+10)^3}}} Cube both sides


{{{y = (-x+10)^3+8}}} Add 8 to both sides


The inverse is {{{g(x) = (-x+10)^3+8}}}


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Let's use the inverse to compute the inputs x = 10, x = 11, x = 12


{{{g(x) = (-x+10)^3+8}}}


{{{g(10) = (-10+10)^3+8}}}


{{{g(10) = (0)^3+8}}}


{{{g(10) = 0+8}}}


{{{g(10) = 8}}}


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{{{g(x) = (-x+10)^3+8}}}


{{{g(11) = (-11+10)^3+8}}}


{{{g(11) = (-1)^3+8}}}


{{{g(11) = -1+8}}}


{{{g(11) = 7}}}


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{{{g(x) = (x+10)^3+8}}}


{{{g(12) = (-12+10)^3+8}}}


{{{g(12) = (-2)^3+8}}}


{{{g(12) = -8+8}}}


{{{g(12) = 0}}}
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