# Prove a function is strictly positive on an interval

How can I prove the function $f(t) = t\left(s-1\right)+1-s^t$ is strictly positive for $t \in (0, 1)$ when $s \geq 0, s \neq 1$? I tried using inequalities but could not find a satisfactory solution.

Here is a plot of the relevant graph (in green) on Desmos: https://www.desmos.com/calculator/wqttkdfinw.

• presumably $s$ is more constrained than that? for example, I would think $s$ is positive – qbert Apr 20 '18 at 22:37
• @qbert Correct. – user76284 Apr 20 '18 at 22:38

Hint:   the case $s=0$ is trivial, otherwise note that $\,f(0)=f(1)=0\,$ and $f$ is strictly concave for $s \gt 0, s \ne 1$ since $f''(t) = - \ln^2(s) \cdot s^t \lt 0\,$.
• what is baffling with this solution is that $f''$ is such easy to study, while both $f$ and $f'$ are intractable. – zwim Apr 20 '18 at 23:32
• @zwim The problem sort of obfuscates the simpler fact that $f(t)=0$ are the intersections between the (convex) exponential $s^t$ and the line $(s-1)t+1$, which so "happen" to be the endpoints of $(0,1)$. – dxiv Apr 20 '18 at 23:48
• I'm probably being dense, but how does this show $f(t) > 0$? I can "see" why it's the case intuitively, especially when I plot the graph, but I'm having trouble formally proving it. What are the relevant theorems? Intermediate value theorem? – user76284 Apr 21 '18 at 0:02
• @user76284 Jensen's inequality: $\;f(t) = f\big((1-t)\cdot 0 + t \cdot 1\big) \gt (1-t)\cdot f(0) + t \cdot f(1) = 0\,$. – dxiv Apr 21 '18 at 0:15