Problem 34 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ f(x)=x^{5}+x^{3}+2 x-1 $$

Step-by-Step Solution

Answer

The function is increasing on all real numbers because its derivative is always positive.

Step 1: Find the first derivative

To determine when the function is increasing or decreasing, we first need to find its derivative. The function given is \( f(x) = x^{5} + x^{3} + 2x - 1 \). We will differentiate each term: \[ f'(x) = 5x^{4} + 3x^{2} + 2. \]

Step 2: Determine critical points

Critical points occur where the derivative is zero or undefined. The derivative here is a polynomial, hence never undefined. Set the derivative equal to zero and solve: \[ 5x^{4} + 3x^{2} + 2 = 0. \] This equation cannot be factored easily, so we typically use numerical methods or graphing to approximate the roots. However, we can assert that it's generally non-zero for all real \(x\) because the leading term \(5x^4\) dominates, causing it to essentially be always positive.

Step 3: Analyze the sign of the derivative

Since \(5x^{4} + 3x^{2} + 2\) is a polynomial with even degree and positive lead coefficient, it does not cross zero unless there's an error in algebra. Its nature implies the derivative is always positive for any real \(x\), confirming \(f(x)\) is increasing everywhere on its domain.

Key Concepts

Increasing and Decreasing Functions

Understanding where a function increases or decreases helps us understand its behavior.
A function is said to be "increasing" when, as you move from left to right on a graph, the function values grow. Conversely, a function is "decreasing" when the values drop as you go from left to right.
You can determine these intervals by examining the first derivative of a function.

• If the derivative is positive over an interval, the function increases on that interval.
• If the derivative is negative, it decreases.

In the given exercise, since the derivative is positive everywhere, the function increases on its entire domain.

First Derivative

The first derivative of a function is a critical tool in calculus because it tells us about the function's rate of change. In simple terms, it helps us know whether a function is moving upwards or downwards.
In calculus, finding the first derivative often involves differentiating each term in the function separately. For polynomial functions like the one given in the exercise, this involves applying basic derivative rules:

• If given a term like \(x^n\), the derivative is \(nx^{n-1}\).

The derivative of the function \(f(x) = x^{5} + x^{3} + 2x - 1\) is \(f'(x) = 5x^{4} + 3x^{2} + 2\). This derivative helps to identify regions where the function increases or decreases.

Polynomial Functions

Polynomial functions are sums of terms consisting of a variable raised to a power and multiplied by a coefficient. They are versatile and can model various behaviors.

• They can have multiple turning points, known as local maxima and minima.
• The degree of the polynomial determines its general shape – a fifth-degree polynomial, like our example, can have up to five zeroes or roots.

Because derivatives of polynomials are also polynomials, they are straightforward to differentiate.
This makes working with polynomial functions relatively easy, especially when identifying when they might increase or decrease.

Critical Points

Critical points are where important changes in the function's behavior, like peaks or valleys, occur. To find them, set the first derivative to zero and solve for the variable.
At these points:

• The function's rate of change reaches zero.
• They can indicate local maxima, local minima, or points of inflection.

In the quadratic highlighted by the exercise, the derivative \(5x^{4} + 3x^{2} + 2\) does not zero out, meaning in this circumstance, no critical points denote a change from increasing to decreasing. This unique case tells us that the function is steadily increasing across its entire domain, as verified by the consistently positive derivative.