Solve Multivariable Sketch Problem: Power of e Function

[seiferseph]

Here is the problem I'm having some trouble with. The answer is fairly simple, it is the power of the e function. (the parabola x = y^2 + 1) I'm not sure how to get that, i could use some hints/help, thanks!

 

[StatusX]

Think about what values for the power of an exponent give a result greater than one.

 

[HallsofIvy]

Are you aware that "The answer is fairly simple, it is the power of the e function. (the parabola x = y^2 + 1)" is NOT precisely true? That is, the region you are asked to sketch is NOT the parabola itself. The parabola is a boundary of the region. Indeed if you think about the fact that "power of the e function" is NOT x= y^2+ 1 nor a parabola, you might see what StatusX means!

 

[benorin]

I'll be kind: recall that $$e^w \geq 1$$ if, and only if, $$w\geq 0,$$ that is, if, and only if the exponent of e is greater than or equal to 0. You are asked to find (and sketch) all points in xy-plane (e.g. all values of x and y) such that $$e^{1-x+y^{2}}\geq 1,$$ which occurs if, and only if, $$1-x+y^{2}\geq 0$$ which describes the region in the xy-plane bounded by (and to the left of) the parabola $$x=1+y^{2}$$.

 

[seiferseph]

I'll be kind: recall that $$e^w \geq 1$$ if, and only if, $$w\geq 0,$$ that is, if, and only if the exponent of e is greater than or equal to 0. You are asked to find (and sketch) all points in xy-plane (e.g. all values of x and y) such that $$e^{1-x+y^{2}}\geq 1,$$ which occurs if, and only if, $$1-x+y^{2}\geq 0$$ which describes the region in the xy-plane bounded by (and to the left of) the parabola $$x=1+y^{2}$$.

ohh, i see, i forgot that to get that's how to get the entire function greater than 1, i get it now, thanks everyone!