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Chapter 4: Problem 5

Find all real numbers (if any) that are fixed points for the given functions. $$h(x)=x^{2}-3 x-5$$

Short Answer

Expert verified
The fixed points are \( x = 5 \) and \( x = -1 \).

Step by step solution

01

Understanding the Definition of a Fixed Point

A fixed point of a function \( h(x) \) occurs when \( h(x) = x \). This means that the point \( x \) does not change the value of the function.
02

Setting Up the Equation

To find the fixed points, we set \( h(x) = x \). Therefore, we have the equation \( x^2 - 3x - 5 = x \).
03

Rearranging the Equation

Rearrange the equation to have all terms on one side, resulting in: \( x^2 - 3x - 5 - x = 0 \). Simplify this to obtain \( x^2 - 4x - 5 = 0 \).
04

Solving the Quadratic Equation

The equation \( x^2 - 4x - 5 = 0 \) is a standard quadratic equation. To solve it, we can use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -4 \), and \( c = -5 \).
05

Calculating the Discriminant

Calculate the discriminant: \( b^2 - 4ac = (-4)^2 - 4(1)(-5) = 16 + 20 = 36 \). Since the discriminant is positive, there are two real solutions.
06

Applying the Quadratic Formula

Use the quadratic formula to find the roots: \( x = \frac{-(-4) \pm \sqrt{36}}{2(1)} = \frac{4 \pm 6}{2} \).
07

Determining the Solutions

Calculate the two possible solutions: \( x = \frac{4 + 6}{2} = 5 \) and \( x = \frac{4 - 6}{2} = -1 \).
08

Verifying the Fixed Points

Verify that at \( x = 5 \), \( h(5) = 5 \). Likewise, check \( x = -1 \), \( h(-1) = -1 \). Both values satisfy the equation \( h(x) = x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Equation
A quadratic equation is an algebraic expression where the highest exponent of the variable is 2. It takes the general form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) cannot be zero. Understanding quadratics is crucial because they frequently appear in various mathematical contexts, including physics, engineering, and economics.
Quadratic equations can have:
  • Two real solutions
  • One real solution (a repeated root)
  • No real solutions (if solutions are complex)
This type of equation graphically represents a parabola in the Cartesian plane, which can open upwards or downwards depending on the sign of \( a \). In our example, we had \( x^2 - 4x - 5 = 0 \), indicating a parabola where the solutions are the points intercepting the x-axis.
Discriminant
The discriminant is a key concept in determining the nature of the solutions of a quadratic equation. For the quadratic equation \( ax^2 + bx + c = 0 \), the discriminant is given by the formula \( b^2 - 4ac \). The discriminant helps us understand how many real roots our quadratic equation will have.
Here's what the discriminant tells us:
  • If \( b^2 - 4ac > 0 \), there are two distinct real roots.
  • If \( b^2 - 4ac = 0 \), there is exactly one real root, meaning the parabola touches the x-axis at one point (tangent).
  • If \( b^2 - 4ac < 0 \), there are no real roots, as the solutions are complex numbers (the parabola doesn鈥檛 intersect the x-axis).
In our exercise, the discriminant was calculated as 36, which is positive, indicating two distinct real roots.
Quadratic Formula
The quadratic formula is a powerful tool used to find the solutions of a quadratic equation. For the equation \( ax^2 + bx + c = 0 \), the solutions can be found using:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula helps find the roots of any quadratic equation by simply substituting the values of \( a \), \( b \), and \( c \) into it. It鈥檚 universally applicable and especially useful when factoring is complex or impossible.
Steps to use the quadratic formula:
  • Identify coefficients \( a \), \( b \), and \( c \) from the equation.
  • Calculate the discriminant \( b^2 - 4ac \).
  • Substitute \( b \), the discriminant, and \( a \) into the formula.
  • Compute the values to get the solutions.
In the given problem, applying the quadratic formula yielded the solutions \( x = 5 \) and \( x = -1 \), confirming our real number fixed points where the function equals the variable.

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Most popular questions from this chapter

A piece of wire 16 in. long is to be cut into two pieces. Let \(x\) denote the length of the first piece and \(16-x\) the length of the second. The first piece is to be bent into a circle and the second piece into a square. (a) Express the total combined area \(A\) of the circle and the square as a function of \(x\) (b) For which value of \(x\) is the area \(A\) a minimum? (c) Using the \(x\) -value that you found in part (b), find the ratio of the lengths of the shorter to the longer piece of wire.

(a) Use a graphing utility to draw a graph of each function. (b) For each \(x\) -intercept, zoom in until you can estimate it accurately to the nearest one-tenth. (c) Use algebra to determine each \(x\) -intercept. If an intercept involves a radical, give that answer as well as a calculator approximation rounded to three decimal places. Check to see that your results are consistent with the graphical estimates obtained in part (b). $$N(t)=t^{7}+8 t^{4}+16 t$$

A wire of length \(L\) is cut into two pieces. The first piece is bent into a square, the second into an equilateral triangle. Express the combined total area of the square and the triangle as a function of \(x,\) where \(x\) denotes the length of wire used for the triangle. (Here, \(L\) is a constant, not another variable.)

Find quadratic functions satisfying the given conditions. The vertex is \((3,-1),\) and one \(x\) -intercept is 1.

If \(f(x)=a x^{2}+b x+c,\) show that \(\frac{f(x+h)-f(x)}{h}=2 a x+a h+b\).
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