y=exxy = e^x - x

Graph of the Function y=exxy = e^x - x

y=exxy = e^x - x is an exponential with a linear function subtracted. The fact that its minimum value is exactly 11 is precisely the most basic inequality about the exponential function, exx+1e^x \geq x + 1.

Domain and monotonicity

The domain is all real numbers. The derivative is as follows.

y=ex1y' = e^x - 1

It vanishes when ex=1e^x = 1, that is at x=0x = 0. It is negative for x<0x < 0, where ex<1e^x < 1, and positive for x>0x > 0, so the function has a minimum, both local and global, at x=0x = 0, of value e00=1e^{0} - 0 = 1.

Range and the fundamental inequality

Since the minimum is 11, the range is y1y \geq 1. Rewriting this gives the following, valid for every real xx.

exx+1e^{x} \geq x + 1

Equality holds only at x=0x = 0. The line y=x+1y = x + 1 is the tangent to y=exy = e^{x} at x=0x = 0, so the inequality is an instance of the fact that a concave-up curve lies above its tangent lines. The graph of y=exxy = e^x - x can be read as the vertical gap between those two.

Concavity

The second derivative is y=exy'' = e^{x}, always positive, so the curve is concave up everywhere and has no inflection point. That convexity is exactly why the tangent-line inequality holds.

Asymptotes

As xx \to -\infty the term exe^{x} tends to 00 and yy approaches x-x; indeed y(x)=ex0y - (-x) = e^{x} \to 0, so the line y=xy = -x is a slant asymptote. As x+x \to +\infty, by contrast, exe^{x} overwhelms xx and the function diverges with no asymptote at all. The result is a markedly asymmetric shape: hugging a straight line on the left, shooting up exponentially on the right.

Counting the solutions of an equation

Because y1>0y \geq 1 > 0 there is no xx-intercept, so the equation ex=xe^{x} = x has no real solution. More generally the number of solutions of ex=x+ce^{x} = x + c is the number of intersections of this graph with the horizontal line y=cy = c.

Range of ccNumber of solutions
c<1c < 1none
c=1c = 111, at x=0x = 0
c>1c > 122

Consequences

Substituting lnt\ln t for xx gives an inequality about the logarithm.

lntt1(t>0)\ln t \leq t - 1 \quad (t > 0)

Bounding a logarithm above by a linear function in this way is the starting point of many proofs, including the relation between the arithmetic and geometric means and Gibbs' inequality in information theory. Of all the inequalities about the exponential function, exx+1e^{x} \geq x + 1 is the one to learn first.

Sample points

The curve passes through (0,1)(0, 1), (1,e1)(1,1.718)(1, e - 1) \approx (1, 1.718), (1,e1+1)(1,1.368)(-1, e^{-1} + 1) \approx (-1, 1.368) and (2,e22)(2,5.389)(2, e^{2} - 2) \approx (2, 5.389).