Monotonicity Based Problems - Inequality Based

Monotonicity in mathematics refers to the behavior of a function as it either consistently increases or decreases. Understanding monotonicity is crucial when dealing with inequality-based problems because it helps in determining the intervals where a function is either above or below a certain value, which is essential in solving inequalities.

Understanding Monotonicity

A function $f(x)$ is said to be:

• Monotonically Increasing if for any two points $x_1$ and $x_2$ such that $x_1 < x_2$, we have $f(x_1) \leq f(x_2)$.
• Strictly Monotonically Increasing if for any two points $x_1$ and $x_2$ such that $x_1 < x_2$, we have $f(x_1) < f(x_2)$.
• Monotonically Decreasing if for any two points $x_1$ and $x_2$ such that $x_1 < x_2$, we have $f(x_1) \geq f(x_2)$.
• Strictly Monotonically Decreasing if for any two points $x_1$ and $x_2$ such that $x_1 < x_2$, we have $f(x_1) > f(x_2)$.

Determining Monotonicity

To determine the monotonicity of a function, we use its derivative $f'(x)$:

• If $f'(x) > 0$ for all $x$ in an interval, then $f(x)$ is strictly increasing in that interval.
• If $f'(x) < 0$ for all $x$ in an interval, then $f(x)$ is strictly decreasing in that interval.
• If $f'(x) = 0$ for all $x$ in an interval, then $f(x)$ is constant in that interval.

Monotonicity and Inequalities

When solving inequality-based problems, we often need to find the range of values for which a function is greater than or less than a certain value. Monotonicity can help us determine these ranges by analyzing the intervals of increase and decrease.

Example Problem

Let's consider the function $f(x) = x^3 - 3x + 1$ and solve the inequality $f(x) > 0$.

Step 1: Find the Derivative

First, we find the derivative of $f(x)$:

$$ f'(x) = 3x^2 - 3 $$

Step 2: Determine Critical Points

Next, we find the critical points by setting the derivative equal to zero:

$$ 3x^2 - 3 = 0 $$

$$ x^2 = 1 $$

$$ x = \pm 1 $$

Step 3: Analyze Monotonicity

We create a sign chart to determine the intervals where $f'(x)$ is positive or negative:

Interval	Test Point	$f'(x)$	Monotonicity of $f(x)$
$(-\infty, -1)$	$-2$	$3(4) - 3 = 9$	Increasing
$(-1, 1)$	$0$	$-3$	Decreasing
$(1, \infty)$	$2$	$3(4) - 3 = 9$	Increasing

Step 4: Solve the Inequality

Using the monotonicity information, we can determine where $f(x) > 0$:

• For $x < -1$, $f(x)$ is increasing. We need to check the value at $x = -1$ to see if $f(x)$ is positive or negative to the left of $-1$.
• For $-1 < x < 1$, $f(x)$ is decreasing. We need to check the value at $x = 1$ to see if $f(x)$ is positive or negative to the right of $-1$ and to the left of $1$.
• For $x > 1$, $f(x)$ is increasing. We need to check the value at $x = 1$ to see if $f(x)$ is positive or negative to the right of $1$.

By evaluating $f(-1)$ and $f(1)$, we find:

$$ f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3 > 0 $$ $$ f(1) = (1)^3 - 3(1) + 1 = 1 - 3 + 1 = -1 < 0 $$

Therefore, $f(x) > 0$ for $x < -1$ and $x > 1$.

Conclusion

The solution to the inequality $f(x) > 0$ is:

$$ x \in (-\infty, -1) \cup (1, \infty) $$

Summary Table

Property	Monotonically Increasing	Monotonically Decreasing
Definition	$f(x_1) \leq f(x_2)$ for $x_1 < x_2$	$f(x_1) \geq f(x_2)$ for $x_1 < x_2$
Strict Definition	$f(x_1) < f(x_2)$ for $x_1 < x_2$	$f(x_1) > f(x_2)$ for $x_1 < x_2$
Derivative $f'(x)$	$f'(x) \geq 0$	$f'(x) \leq 0$
Strict Derivative	$f'(x) > 0$	$f'(x) < 0$

By understanding and applying the concept of monotonicity, we can effectively solve inequality-based problems in calculus and other areas of mathematics.