General guidelines to solve Mean Value Theorem problems I am wondering if there is a general guideline to solve this specific type of MVT problem. For the teachers: how do you explain to the students how to apply MVT for these two questions below?
Question #1

Using $MVT$ prove that
$$ e^{x} > 1+x $$
for all $x > 0$

Question #2

Using $MVT$ prove that
$$\ln (1+x) < x$$
for all $x > 0$

For example, when dealing with 

Show that
$$  x^3+e^x=0 $$
cannot have two zeros

I find helpful to 


*

*Find values for the function where $f_{x_1} < 0$ and $f_{x_2} > 0$, in this case for example $x_1 = -2$ and $x_2 = 1$

*Show using the derivative that f'(x) is always positive (or negative, in another case)
Do you have guidelines for the first two questions introduced?
 A: From both your questions it appears that you want to use MVT to establish inequalities of the form $f(x) > g(x)$ for some range of values of $x$ (actually the values of $x$ form an interval $I$).
And MVT is a good tool to prove such inequalities. Let the interval $I$ be of the form $(a, \infty)$ so that we need to prove the inequality $f(x) > g(x)$ for all $x > a$. We need to assume that $f, g$ are continuous on $[a, \infty)$ and differentiable on $(a, \infty)$. Consider $h(x) = f(x) - g(x)$ then we want to have $h(a) \geq 0$ otherwise we can't proceed with this approach. So we assume that $h(a) \geq 0$ so that $f(a) \geq g(a)$. Now by mean value theorem if $x > a$ then $$h(x) - h(a) = (x - a)h'(c)$$ for some $c \in (a, x)$. If $h'(x) > 0$ for all $x > a$ then $h'(c) > 0$ and therefore $h(x) - h(a) > 0$ and hence $h(x) > h(a) \geq 0$ so that $h(x) > 0$ for all $x > a$. You can choose $h(x) = e^{x} - x - 1$ for first problem and $h(x) = log(1 + x) - x$ in second problem.
BTW we don't use MVT explicitly in such problems rather we rely on the following consequences of MVT:


*

*If derivative of a function in positive in an open interval then the function is strictly increasing on the corresponding closed interval.

*If derivative of a function is negative in an open interval then the function is strictly decreasing on the corresponding closed interval.



For the problems you have given there is slightly easier approach which uses MVT in a very direct manner. Since $x > 0$ for both problems we can see that the desired inequalities can be written in the form $$\frac{e^{x} - e^{0}}{x - 0} > 1, \frac{\log(1 + x) - \log(1 + 0)}{x - 0} < 1$$ and the LHS of both inequalities looks like a difference quotient and hence by MVT can be expressed as $f'(c)$ for appropriate $f$ and $c$. Thus for first inequality we have $f(x) = e^{x}$ and $0 < c < x$ and hence the inequality is equivalent to $f'(c) > 1$ or $e^{c} > 1$ which is true because $c > 0$. For second inequality we need $f(x) = \log(1 + x)$.

The third question is more concerned with Rolle's Theorem. If a function has two distinct zeroes then there is a zero of the derivative between these zeroes of original function. Thus if we can show that the derivative has no zeroes then we can ensure that the original function can't have two zeroes. For $f(x) = e^{x} + x^{3}$ we have $f'(x) = e^{x} + 3x^{2} > 0$ and derivative never vanishes and hence $f$ can't have two zeroes.
A: In question 1, you can define the function $ f(x) = e^x - x - 1 $ and note that you have
$$ \frac{f(x) - f(0)}{x - 0} = f'(t) = e^t - 1 $$
for some $ t \in [0, x] $ by the mean value theorem. Dealing with the cases $ x > 0 $ and $ x < 0 $ separately then yields your inequality. The second question is the same inequality; just use monotonicity of the natural logarithm when taking logs on both sides.
A: for the first question you can take $f(t)=e^t$ and the desired interval $[0,x]$ then use MVT. For the second question set $f(t)=ln(1+t)$ and interval $[0,x]$. For the third question$f(x)=x^3+e^x$ use Bolzano theorem for proving the existence of a root $(f(-1)f(0)<0)$ then use Roll's theorem which leads to the contradiction $3x^2+e^x=0$.
