Question 1021098
{{{y=-x^2}}}
To find the inverse interchange x and y and solve for the new y.
The new y is the inverse.
{{{x=-y^2}}}
{{{y^2=-x}}}
{{{y=0 +- sqrt(-x)}}}
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{{{-x^2=-sqrt(-x)}}}
{{{x^2=sqrt(-x)}}}
{{{x^4=-x}}}
{{{x^4+x=0}}}
{{{x(x^3+1)=0}}}
{{{x(x+1)(x^2-x+1)=0}}}
Two solutions:
{{{x=0}}} Not in the range we're interested in.
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{{{x+1=0}}}
{{{x=-1}}}
 *[illustration dx8.JPG].