Problem 56 - Calculus Early Transcendentals

Question

Give a graph of the polynomial and label the coordinates of the intercepts, stationary points, and inflection points. Check your work with a graphing utility. $$ p(x)=x^{4}-6 x^{2}+5 $$

Step-by-Step Solution

Answer

Intercepts: \((0,5), (\pm\sqrt{5}, 0), (\pm 1, 0)\). Stationary points: \((0,5), (\pm\sqrt{3}, -4)\). Inflection points: \((\pm 1, 0)\).

Step 1: Determine Intercepts

To find the x-intercepts, solve \( p(x) = 0 \): \[x^4 - 6x^2 + 5 = 0\]By substitution, let \( y = x^2 \). Then the equation becomes \( y^2 - 6y + 5 = 0 \). Solving this quadratic equation using the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[y = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}\]Thus, \( y = 5 \) or \( y = 1 \). Substituting back, we find \( x^2 = 5 \) or \( x^2 = 1 \), giving the solutions \( x = \pm \sqrt{5} \) and \( x = \pm 1 \).The y-intercept is the value of \( p(0) \), which is given by setting all \( x \) terms to 0. Thus, \( p(0) = 5 \). The intercepts are \( (\pm \sqrt{5}, 0) \), \( (\pm 1, 0) \), and \( (0, 5) \).

Step 2: Identify Stationary Points

Stationary points occur where the derivative is zero. Find the derivative \( p'(x) \): \[p'(x) = \frac{d}{dx}(x^4 - 6x^2 + 5) = 4x^3 - 12x\]Set \( p'(x) = 0 \):\[4x^3 - 12x = 0\]Factor out \( 4x \):\[4x(x^2 - 3) = 0\]Solve for \( x \): \( x = 0 \) or \( x^2 = 3 \), giving \( x = \pm \sqrt{3} \).Thus, the stationary points are at \( x = 0 \), \( x = \pm \sqrt{3} \). Evaluate \( p(x) \) at these points: - \( p(0) = 5 \)- \( p(\sqrt{3}) = (\sqrt{3})^4 - 6(\sqrt{3})^2 + 5 = 9 - 18 + 5 = -4 \)- \( p(-\sqrt{3}) = (-\sqrt{3})^4 - 6(-\sqrt{3})^2 + 5 = 9 - 18 + 5 = -4 \)The stationary points are \((0, 5), (\sqrt{3}, -4), (-\sqrt{3}, -4)\).

Step 3: Identify Inflection Points

Inflection points occur where the second derivative is zero and changes sign. Find the second derivative \( p''(x) \): \[p''(x) = \frac{d}{dx}(4x^3 - 12x) = 12x^2 - 12\]Set \( p''(x) = 0 \):\[12x^2 - 12 = 0\]Solve for \( x \):\[x^2 = 1 \quad \Rightarrow \quad x = \pm 1\]Evaluate \( p(x) \) at these points to confirm they are inflection points:- \( p(1) = 1^4 - 6 \times 1^2 + 5 = 0 \)- \( p(-1) = (-1)^4 - 6 \times (-1)^2 + 5 = 0 \)Thus, the inflection points are \((1, 0), (-1, 0)\).

Step 4: Graph the Polynomial

Using the intercepts, stationary points, and inflection points, graph the polynomial \( p(x) = x^4 - 6x^2 + 5 \). Indicate all intercepts, stationary points, and inflection points:- **x-intercepts**: \( (\pm \sqrt{5}, 0) \) and \( (\pm 1, 0) \)- **y-intercept**: \( (0, 5) \)- **Stationary Points**: \((0, 5), (\sqrt{3}, -4), (-\sqrt{3}, -4)\)- **Inflection Points**: \((1, 0), (-1, 0)\)Verify the graph with a graphing utility.

Key Concepts

Polynomial Functions

Polynomial functions are expressions composed of variables and coefficients using only addition, subtraction, multiplication, and non-negative whole number exponents. For example, the polynomial function given in the exercise is \[ p(x) = x^4 - 6x^2 + 5 \].
Polynomials are classified by their degree, which is the highest power of the variable present. Here, the degree is 4, making it a quartic polynomial.
The shape of a polynomial graph is determined by its degree and leading coefficient. Quartic polynomials can have up to three turning points. This results in a wave-like curve that can shift directions multiple times.
Polynomial functions are essential in calculus and algebra because they set the groundwork for more complex expressions and functions.
A polynomial graph is smooth and continuous without any breaks or corners. This quality is crucial in many real-world applications where polynomials model smooth natural processes.

Intercepts in Calculus

Intercepts are the points at which a graph intersects an axis.

• X-intercepts are found by solving the equation where the polynomial equals zero. These are also known as the roots or zeros of the function. In this exercise, the x-intercepts are \((\pm \sqrt{5}, 0)\) and \((\pm 1, 0)\).
• Y-intercept is the point where the graph crosses the y-axis. It is found by evaluating the polynomial function at zero. Here, the y-intercept is \((0, 5)\).

Intercepts are pivotal for constructing and understanding the visual representation of polynomial graphs. They help in identifying intervals of positive or negative values in a function.
In calculus, intercepts are also used to study the behavior of functions near these critical points.

Stationary Points

Stationary points occur where the derivative of the function equals zero, indicating a potential maximum, minimum, or saddle point.

• To find these points, compute the derivative of the polynomial and set it to zero.
  Here, the first derivative is \[ p'(x) = 4x^3 - 12x \].
• Setting this equation to zero, \[ 4x(x^2 - 3) = 0 \], gives solutions at \(x = 0\) and \(x = \pm \sqrt{3}\).
• The corresponding y-values can be found by substituting these x-values back into the original polynomial, providing stationary points.

Stationary points are key in understanding the overall "shape" of the graph because they show where the curve changes.
In optimization problems, these points often indicate optimal solutions where maximum or minimum values are found.

Inflection Points

Inflection points are where the graph of the polynomial changes concavity. That is, it switches from "bending up" to "bending down," or vice versa.

• To find them, calculate the second derivative of the polynomial and solve for zero. In this context, \[ p''(x) = 12x^2 - 12 \].
• Setting this to zero, \[ 12x^2 - 12 = 0 \], reveals inflection points at \(x = \pm 1\).
• Evaluate these points in the original polynomial to confirm they are true inflection points, resulting in points \((1, 0)\) and \((-1, 0)\).

Recognizing inflection points is important in calculus as they provide insights into the "flex" in the graph.
This helps in understanding the behavior of functions across their continua and in various applications such as economics, where shifts in trends are important.