Mean Value Theorem and Inequalities
Lecture 14: Mean Value Theorem and Inequalities
Mean Value Theorem (MVT)
Definition: The Mean Value Theorem states:
If a function $f$ is differentiable on the interval $(a, b)$ and continuous on the closed interval $[a, b]$, then there exists a point $c$ in the interval $(a, b)$ such that:
rac{f(b) - f(a)}{b - a} = f'(c)Here:
rac{f(b) - f(a)}{b - a} represents the slope of the secant line between the points $(a, f(a))$ and $(b, f(b))$.
$f'(c)$ is the slope of the tangent line to the function at the point $c$.
Geometric Proof
Outline of Proof:
Consider drawing lines parallel to the secant line (illustrated in Figure 1).
Move these lines up (or down) from the graph of the function until one line touches the graph.
If the function is not differentiable, this method fails, for instance, the function $f(x) = |x|$ has $f'(0)$ undefined but always touches at $x = 0$. (See Figure 2).
Interpretation of the Mean Value Theorem
Analogy: If you travel from Boston to Chicago (1000 miles in 3 hours), then at some point, your speed must have been $\frac{1000}{3} \text{ mph}$.
This can be formalized as:
Let $f(t) = ext{position from Boston}$ where:
$f(3) = 1000$ miles
$f(0) = 0$ miles
rac{1000 - 0}{3 - 0} = f'(c) where $c$ is the time at which this speed occurred.
Versions of the Mean Value Theorem
Alternate Expressions:
f(b) - f(a) = f'(c)(b - a)
f(b) = f(a) + f'(c)(b - a), for some $c$ where $a < c < b$.
Change $b$ to $x$ results in:
f(x) = f(a) + f'(c)(x - a), for some $c$ where $a < c < x$.
Significance: The different formulations of the MVT highlight its versatility. The exact location of $c$ is not specified and depends on $f$, $a$, and $x$.
Comparison to Linear Approximation
Linear Approximation Formula:
f(x) \approx f(a) + f'(a)(x - a), when $x$ is near $a$.
Difference: The slope in linear approximation ($f'(a)$) is fixed, while the MVT specifies a variable slope ($f'(c)$).
Key Conclusions from the Mean Value Theorem
If $f'(x) > 0$, then $f$ is increasing over the interval.
If $f'(x) < 0$, then $f$ is decreasing over the interval.
If $f'(x) = 0$ for all $x$, then $f$ is constant.
Definitions:
Increasing Function: $a < b \Rightarrow f(a) < f(b)$.
Decreasing Function: $a < b \Rightarrow f(a) > f(b)$.
Proofs of Key Conclusions
Proof of Conclusion 1:
Assume $a < b$.
From the Mean Value Theorem: f(b) = f(a) + f'(c)(b - a)
Both $f'(c) > 0$ and $(b - a) > 0$ imply:
f(b) > f(a).
Proof of Conclusion 2: Similar to Conclusion 1, omitted as it follows the same reasoning.
Proof of Conclusion 3:
Assume:
f(b) = f(a) + f'(c)(b - a)
If $f'(c) = 0$, then:
f(b) = f(a) + 0 imes (b - a) = f(a), establishing constancy.
Understanding the Implications
The logic behind these conclusions is not as trivial as it might appear, requiring an understanding of derivatives and their behavior at infinitesimal scales.
Inequalities Derived from the Mean Value Theorem
The fundamental property $f' > 0 \Rightarrow f$ is increasing leads to several inequalities.
Example Inequalities:
$e^x > 0$ for all $x$.
$e^x > 1$ for $x > 0$.
$e^x > 1 + x$ for $x > 0$.
Proof of Inequalities
Proof of Inequality 2:
Define the function $f_1(x) = e^x - 1$. Therefore:
$f_1(0) = e^0 - 1 = 0$.
f_1'(x) = e^x > 0 \text{ for } x > 0.
This means that $f_1(x)$ is increasing, hence:
f1(x) > f1(0) or e^x > 1 \text{ for } x > 0.
Proof of Inequality 3:
Define the function $f_2(x) = e^x - (1 + x)$. Then:
f2'(x) = e^x - 1 > 0 \text{ for } x > 0 implying that $f2(x)$ is increasing, hence:
f2(0) = 0 \Rightarrow f2(x) > 0 \text{ for } x > 0.
Therefore, we conclude $e^x > 1 + x$ for $x > 0$.
Further Inequalities:
Higher order expansions also yield:
e^x > 1 + x + \frac{x^2}{2!} \text{ for } x > 0.
Continuing this pattern results in:
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots, as discussed in connection with Taylor series towards the end of the course.