Question 1197339
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Graph out y = 6x^2+11 to find that it fails the horizontal line test. 
{{{drawing(500,500,-3,3,-3,15,
graph(500,500,-3,3,-3,15,0,6x^2+11)
)}}}
It is possible to draw a single horizontal line through more than one point on the parabola.
This function is not one-to-one, i.e. the function is not injective.


As such, no inverse exists if we account for the entire domain of f(x) = 6x^2+11


But if we restrict the domain to {{{x >= 0}}}, then the graph passes the horizontal line test. This is the right hand side of the parabola.


The outline to finding the inverse follows these steps
1) Replace f(x) with y
2) Swap x and y
3) Solve for y


Let's follow that process
f(x) = 6x^2+11
y = 6x^2+11
x = 6y^2+11
x-11 = 6y^2
6y^2 = x-11
y^2 = (x-11)/6
y = sqrt( (x-11)/6 ) 



The inverse is *[tex \Large f^{-1}(\text{x}) = \sqrt{\frac{\text{x}-11}{6}}]
{{{drawing(500,500,-3,15,-3,15,
graph(500,500,-3,15,-3,15,0,(sqrt(x)/sqrt(x))*(6x^2+11),sqrt((x-11)/6),(sqrt(sin(5x))/sqrt(sin(5x)))*x)
)}}}
The green graph represents the original function f(x) = 6x^2+11 but only when {{{x >= 0}}}
The inverse is the blue curve, which is a reflection of the green curve over the dashed purple line y = x
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