User:Keith.U/Sandbox/Proof 1

Theorem
Let $a \in \R$ be a real number such that $a < 1$.

Let $x \in \R_{>0}$ be a strictly positive real number such that $0 < x < 1$.

Then:
 * $1 < a^x < 1 + a x$

Proof
Define a real function $g_x: \R_{> 0} \to \R$ as:
 * $g_x \left({a}\right) = 1 + a x - a^x$

Then differentiating $a$ gives:
 * $D_a g_x \left({a}\right) = x \left({1 - a^{x-1} }\right)$

We show now that the derivative of $g_x$ is positive for all $a > 1$:

So $D_a g_x \left({a}\right)$ is positive for all $a > 1$.

Whence, by Derivative of Monotone Function, $g_x$ is increasing for all $a > 1$.

Now, $g_x \left({1}\right) = x > 0$.

So $g_x \left({a}\right)$ is positive for all $a > 1$.

That is:
 * $1 + a x - a^x > 0$

Adding $a^x$ to both sides of the above yields:
 * $a^x < 1 + a x$

Finally:

So, for $0 < x < 1$:
 * $1 < a^x < 1 + a x$