Problem.
Show that $$e^x > 1 + x , \ \ x \neq 0$$
My attempt.
Using Mean Value Theorem: $$f'(c) = \frac{f(b) - f(a)}{b-a}$$
$$\Rightarrow e^c = \frac{e^x - 1}{x}$$ $$\Rightarrow xe^c = e^x -1$$ $$\Rightarrow x+1 = \frac{e^x}{e^c}$$
I'm not feeling very comfortable with these calculations...Am I on the right track? Can I even choose $e^o$ in my initial calculation as it says that $x \neq 0 $ ?
On
Let's consider 2 cases:
a) $x > 0$: Then $(e^x - 1)/x = e^c > 1$ for some $c > 0$ then the conclusion follows.
b) $x < 0$: Then we need to prove: $-x > 1 - e^x$ or $1 > \frac{(1 - e^x)}{(0 - x)}$. Apply MVT again we have $\text{RHS} = e^c$ for some $c \,\text{in}\, (x, 0)$ so $\text{RHS} = e^c < e^0 = 1$ and the conclusion follows.
On
A geometric way to prove the inequality is to use the convexity of the function $f=\exp$ (since its second derivative is $e^x\ge0$) then the line tangent at $x=0$ with equation: $$y=f'(0)x+f(0)=x+1$$ is below the curve of the function hence $$e^x> x+1,\quad x\ne0$$
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Let $$f(x) = e^x - x - 1$$
Therfore you have $$f'(x) = e^x-1 $$
If you have a look at $f'(x)$ you will see that it is always greater than $0$ for $x>0$ . The value of $f(0)=0$ . After that the function is monotonically increasing and will always be $>0$ . Therefore from the above we can conculde that for $x>0$ , $f(x)>0$ . Hence $e^x-x-1>0$ , $e^x>x+1$
On
(1) Assume $x>0$. By MVT, there is a $c \in (0,x)$ so that $e^c = \frac{e^x - 1}{x}$. Then $e^x = xe^c + 1$. But since $c \in (0,x)$, we have $e^c > 1$, and $e^x = xe^c + 1 > x+1$.
(1) Assume $x<0$. By MVT, there is a $c \in (x,0)$ so that $e^c = \frac{1-e^x}{-x}$. Then $e^x = xe^c + 1$. But since $c \in (x,0)$, we have $0 < e^c < 1$, and $e^x = xe^c + 1 > x+1$ (since $x<0$).
On
Let $f$ be the exponential function.
Suppose $x > 0$. Then by the mean-value theorem, there exists some $c$, with $0 < c < x$, such that $$f'(c) = \dfrac{f(x) - f(0)}{x - 0}$$ $$e^c = \dfrac{e^x - 1}{x}.$$ As $c > 0$, we have $e^c > 1$, and therefore $\dfrac{e^x - 1}{x} > 1$. Multiplying both sides by the positive number $x$ gives $e^x - 1 > x$.
Suppose $x < 0$. Then by the mean-value theorem, there exists some $c$, with $x < c < 0$, such that $$f'(c) = \dfrac{f(0) - f(x)}{0 - x}$$ $$e^c = \dfrac{1 - e^x}{-x} = \dfrac{e^x - 1}{x}.$$
As $c < 0$, we have $e^c < 1$, and therefore $\dfrac{e^x - 1}{x} < 1$. Multiplying both sides by the negative number $x$ reverses the inequality, giving $e^x - 1 > x$ again, as required.
On
The Mean Value Theorem is a valid approach. Here is another approach; hopefully, it will make you more comfortable to have two.
The integer version of Bernoulli's Inequality, proven at the end of this answer, is sufficient to prove that, for $n_0\gt\max(-x,1)$ and $x\ne0$, we have $$ 1+x\lt\left(1+\frac x{n_0}\right)^{n_0}\le\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x $$
On
Your approach using the Mean Value Theorem is OK, but you made a mistake in the final line. It would correctly read $x+{1\over e^c}={e^x\over e^c}$ instead of $x+1$ on the left hand side. But you really should have stopped at the second line, $e^c={e^x-1\over x}$ for some $c$ between $0$ and $x$, and then reasoned as follows:
If $x\gt0$, then $c\gt0$. In this case $(e^x-1)/x=e^c\gt1$, hence $e^x-1\gt x$, so that $e^x\gt1+x$.
If $x\lt0$, then $c\lt0$. In this case $(e^x-1)/x=e^c\lt1$, hence (and here you have to remember than multiplying by a negative number reverses the direction of the inequality!) $e^x-1\gt x$, so that $e^x\gt1+x$.
The answer can be solved using the Taylor Series of $e^x$, which is $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$. Note that if $x > 0$, each and every term in the series will be greater than $0$, and therefore $e^x > 1+x$.