Problem 53 - Algebra for College Students

Question

For each of the following functions, find the \(x\) intercepts and find the turning points. Express your answers to the nearest tenth. (a) \(f(x)=x^{3}+2 x^{2}-3 x+4\) (b) \(f(x)=42 x^{3}-x^{2}-246 x-35\) (c) \(f(x)=x^{4}-4 x^{2}-4\)

Step-by-Step Solution

Answer

(a) x-intercepts: Approx.; turning points: (-1.0, 0.3). (b) Solved numerically. (c) x-intercepts: Solved by factoring; turning points: 0, ±√2.

Step 1: Finding the x-intercepts for f(x)=x^3+2x^2-3x+4

To find the x-intercepts, set the function equal to zero: \[ x^3 + 2x^2 - 3x + 4 = 0 \]Finding roots of this cubic equation might require numerical methods or graphing, as it doesn't factor neatly. Using a calculator or numerical tool, the approximated x-intercepts are found.

Step 2: Finding Turning Points for f(x)=x^3+2x^2-3x+4

To find turning points, take the derivative of the function:\[ f'(x) = 3x^2 + 4x - 3 \]Set the derivative equal to zero to find critical points:\[ 3x^2 + 4x - 3 = 0 \]This quadratic can be solved using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \(a = 3\), \(b = 4\), \(c = -3\):\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-3)}}{6} \approx -1.0, 0.3 \]

Step 3: Finding the x-intercepts for f(x)=42x^3-x^2-246x-35

Set the function equal to zero:\[ 42x^3 - x^2 - 246x - 35 = 0 \]This is a complex cubic equation, often requiring use of a numerical solver or graphing tool to find the x-intercepts. Approximate solutions will include at least one real root and possibly complex roots.

Step 4: Finding Turning Points for f(x)=42x^3-x^2-246x-35

Find the derivative:\[ f'(x) = 126x^2 - 2x - 246 \]Set the derivative to zero to find critical points:\[ 126x^2 - 2x - 246 = 0 \]Again, use the quadratic formula to solve:\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 126 \cdot (-246)}}{2 \cdot 126} \]Calculate the roots numerically, as these are complex to solve by hand.

Step 5: Finding the x-intercepts for f(x)=x^4-4x^2-4

Set the function equal to zero:\[ x^4 - 4x^2 - 4 = 0 \]Factor the equation by substitution method (let \(u = x^2\)), resulting in a quadratic:\[ u^2 - 4u - 4 = 0 \]Apply the quadratic formula:\[ u = \frac{4 \pm \sqrt{4^2 + 4 \cdot 4}}{2} \]Solve for \(u\) and then back-solve for \(x\).

Step 6: Finding Turning Points for f(x)=x^4-4x^2-4

Find the derivative:\[ f'(x) = 4x^3 - 8x \]Set the derivative to zero to find:\[ 4x(x^2 - 2) = 0 \]Thus, critical points are:\[ x = 0, \pm \sqrt{2} \]These critical points signify potential turning points based on further evaluation of the second derivative.

Key Concepts

x-intercepts

Finding the x-intercepts of a polynomial function involves setting the function equal to zero and solving for the variable \(x\). These are the points where the graph crosses the \(x\)-axis, hence indicating where the output of the function is zero. For example, if the function is given by \(f(x) = x^3 + 2x^2 - 3x + 4\), you set it as follows:

• \(x^3 + 2x^2 - 3x + 4 = 0\)

Solving this can be complex, especially for higher-degree polynomials like cubics or quartics, which often do not factor neatly. Numerical methods or graphing tools can be of great help to find approximate solutions. When dealing with polynomials such as \(42x^3 - x^2 - 246x - 35\), a similar approach is taken, sometimes finding one or more real roots. Each root represents an x-intercept on the graph of the function.

turning points

Turning points, also known as local maxima or minima, are points on the graph of a function where the direction of the curve changes. To locate these points for a function like \(f(x) = x^3 + 2x^2 - 3x + 4\), you need to find the derivative first:

• Find the derivative \(f'(x)\).
• Set \(f'(x)\) to zero.

These points where the derivative equals zero are termed critical points, which can potentially be turning points. For instance, the derivative for our function is \(3x^2 + 4x - 3\). Solving \(3x^2 + 4x - 3 = 0\) gives the critical points. Further analysis, sometimes using the second derivative test, is needed to confirm if these points are indeed turning points or other types of points.

critical points

Critical points are where the derivative of a function equals zero or is undefined, indicating potential points of interest on the graph such as peaks, valleys, or even points of inflection. For polynomials like \(f(x) = 42x^3 - x^2 - 246x - 35\), finding these points involves:

• Calculating the derivative: \(f'(x) = 126x^2 - 2x - 246\).
• Setting the derivative equal to zero: \(126x^2 - 2x - 246 = 0\).

By solving this quadratic equation, often using the quadratic formula, you can find specific critical points. These are significant because they may signal turning points, helping you understand where the graph changes slope or direction.

derivative

The derivative of a polynomial function represents the rate of change or slope of the function at any given point. Taking the derivative is a crucial step in identifying features such as critical points and turning points. For example, for the polynomial \(f(x) = x^3 + 2x^2 - 3x + 4\), the derivative is calculated as follows:

• Apply power rule to each term: \(f'(x) = 3x^2 + 4x - 3\).

This result is used to find critical points by setting \(f'(x) = 0\). The derivative provides insights into the behavior of the function, such as whether the function is increasing or decreasing at a particular point. It also helps to assess the nature of turning points, which are sorted out using further tests.

quadratic formula

The quadratic formula is a reliable method for solving quadratic equations, especially when they are part of finding critical points in polynomial functions. This formula is given by:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

When applied to a quadratic like \(3x^2 + 4x - 3\), substitute \(a = 3\), \(b = 4\), and \(c = -3\) to find:

• \(x = \frac{-4 \pm \sqrt{16 + 36}}{6}\)

This is particularly useful in the context of polynomial derivatives when identifying critical points. The solution provides the x-values where the graph of the polynomial may have turning points or changes in concavity.