How to show $f(x)=(e^x-1)/x, x>0$ is convex?

Question

How does one show that $f(x)=(e^x-1)/x$ is convex on $(0,\infty)$? I plotted the curve and it looks clearly convex. However, when I tried differentiating it, I cannot show the second derivative, $$ f''(x) = \frac{x^3e^x-2x^2e^x+2xe^x-2x}{x^4},$$ remains nonnegative on $(0,\infty)$.

Try looking at the power series? – Steven Gubkin Jan 25 '16 at 16:42 — Steven Gubkin, Jan 25 '16 at 16:42

score 3 · Accepted Answer · answered Jan 25 '16 at 16:42

HINT

I think the easy way to do this is with Taylor series. Note $$ f(x) = \frac{e^x-1}{x} = \sum_{k=1}^\infty \frac{x^{k-1}}{k!} $$ so $$ f''(x) = \sum_{k=3}^\infty \frac{(k-1)(k-2)}{k!} x^{k-3} $$ where every term is non-negative for non-negative $x$...

score 2 · Answer 2 · edited Jan 25 '16 at 16:50

2

The power series has all positive coefficients, so all derivatives are positive for $x > 0$.

edited Jan 25 '16 at 16:50

imranfat

10,029

answered Jan 25 '16 at 16:43

marty cohen

107,799

How to show $f(x)=(e^x-1)/x, x>0$ is convex?

2 Answers2