Why does e^x grow faster than x^2 for large x?

phospho · Aug 9, 2012

The graph e^x = x^2:

I originally drew it like this:

Hand-drawn sketch of x-y axes with a black curve for e^x and a red curve for x^2; both curves cross near the origin and the e^x curve rises less steeply than x^2 on the right.

But it's actually:

Graph with x and y axes meeting at origin O, showing the parabola y = x^2 passing through O and the decreasing curve y = e^(-x) that approaches the x-axis as x grows and crosses the y-axis above O.

If I come across more complex graphs in the future, is there a way to know which one is steeper than the other, to draw it accurately ?

Jakeus314 · Aug 9, 2012

phospho said:

The graph e^x = x^2:

I originally drew it like this:

But it's actually:

If I come across more complex graphs in the future, is there a way to know which one is steeper than the other, to draw it accurately ?

You could plot some actual points of interest to see where the plots are...

lewando · Aug 9, 2012

A good plotting tool exists at fooplot.com -- by the way, "The graph e^x = x^2:" makes little sense.

Mentallic · Aug 10, 2012

phospho said:

If I come across more complex graphs in the future, is there a way to know which one is steeper than the other, to draw it accurately ?

Yes, they're called derivatives, but you study those in calculus and since you posted this in precalculus I'll leave it at that.

Why does e^x grow faster than x^2 for large x?

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