Value of parameter

This is a very classic exercise and can be dealt with various ways. We know the result in advance. Why? Because it is the Taylor Polynomial of the exponential function. Let us see however how we gonna deal with it with High School Methods.

Find the positive real number $\alpha$ such that

$e^x \geq 1+x+\frac{x^2}{2} + \alpha x^3 \quad \quad \text{forall} \;\; x \in \mathbb{R}$

Solution

Define the function

$f(x) = \frac{1+x+\frac{x^2}{2}+\alpha x^3}{e^x} \quad , \quad x \in \mathbb{R}$

and note that $f(x) \leq 1$ forall $x \in \mathbb{R}$ . Clearly , $f$ is differentiable and its derivative is given by

$f'(x)= \frac{x^2\left ( 6\alpha-1-2\alpha x \right )}{e^x} \quad , \quad x \in \mathbb{R}$

It follows that $\displaystyle f'(x) =0 \Leftrightarrow \left\{\begin{matrix} x &= &0 \\\\ x & = & \dfrac{6\alpha-1}{2\alpha} \end{matrix}\right.$ . Suppose that $\frac{6\alpha-1}{2\alpha}>0$ . Then the monotony of $f$ as well as the sign of $f'$ is seen at the following table.

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It follows then that $\displaystyle f\left ( \frac{6\alpha-1}{2\alpha} \right )> f(0)=1$ . This is an obscurity due to the fact that $f(x) \leq 1$ . Similarly, if we suppose that $\frac{6\alpha-1}{2\alpha}<0$ . Hence

$\frac{6\alpha-1}{2\alpha}=0 \Leftrightarrow \alpha=\frac{1}{6}$

For $\alpha=\frac{1}{6}$ we easily see that the given inequality holds.

Value of parameter

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