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Chapter 1: Problem 37

Given the function \(f(x)=x^{2}-3 x :\) a. Evaluate \(f(5)\) . b. Solve \(f(x)=4\)

Short Answer

Expert verified
f(5) = 10; Solutions to f(x) = 4 are x = 4 and x = -1.

Step by step solution

01

Evaluate f(x) at x = 5

To find the value of the function at a specific point, replace the variable \(x\) in the function \(f(x) = x^2 - 3x\) with the desired value. Here, we need to evaluate \(f(5)\). Substitute \(x = 5\) into the equation:\[ f(5) = 5^2 - 3(5) \] Calculate:- \(5^2 = 25\)- \(-3(5) = -15\)Thus,\[ f(5) = 25 - 15 = 10 \]
02

Set up the equation f(x) = 4

To solve for \(x\) when \(f(x) = 4\), set the function \(f(x) = x^2 - 3x\) equal to 4:\[ x^2 - 3x = 4 \] Rearrange the equation to form a standard quadratic equation:\[ x^2 - 3x - 4 = 0 \]
03

Solve the Quadratic Equation

The equation \(x^2 - 3x - 4 = 0\) is a quadratic equation. We can solve it using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \(a = 1\), \(b = -3\), \(c = -4\). Substitute these values into the formula:\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)} \] \[ x = \frac{3 \pm \sqrt{9 + 16}}{2} \] \[ x = \frac{3 \pm \sqrt{25}}{2} \] \[ x = \frac{3 \pm 5}{2} \] Calculate the two possible values for \(x\):1. \( x = \frac{3 + 5}{2} = 4 \)2. \( x = \frac{3 - 5}{2} = -1 \)Thus, the solutions are \(x = 4\) and \(x = -1\).

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Key Concepts

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

Evaluating Functions
Evaluating a function means finding the value of the function for a specific input or argument. In simpler terms, it's like plugging in a number into the equation and getting a result. Let's look at how we can evaluate a quadratic function like the given one, \( f(x)=x^2-3x \). To evaluate \( f(5) \), we substitute \( x = 5 \) into the function:
  • Calculate \( 5^2 \) which gives us 25.
  • Calculate \(-3 \times 5\), resulting in -15.
Putting it all together: \[ f(5) = 25 - 15 = 10 \]So, when \( x = 5 \), \( f(x) = 10 \). This process can be used for any function, just substitute the given value and solve.
Solving Quadratic Equations
A quadratic equation often takes the form \( ax^2 + bx + c = 0 \). Solving these equations involves finding the values of \( x \) that make the equation true. We are given the quadratic equation \( x^2 - 3x = 4 \). To solve it, we first rearrange it to the standard form: \[ x^2 - 3x - 4 = 0 \] This equation is now ready to be solved using various methods such as factoring, completing the square, or using the quadratic formula.
  • Factoring: Only possible when the equation can be easily broken into simpler binomial expressions.
  • Completing the square: Useful when the equation doesn鈥檛 factor neatly.
  • Quadratic Formula: A reliable method when other methods are cumbersome.
Quadratic Formula
The quadratic formula is a universal tool for solving any quadratic equation. It is particularly useful when factoring is difficult or impossible. The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation \( x^2 - 3x - 4 = 0 \), identify the coefficients:
  • \( a = 1 \)
  • \( b = -3 \)
  • \( c = -4 \)
Substitute these into the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot -4}}{2 \cdot 1} \] This simplifies to: \[ x = \frac{3 \pm 5}{2} \] Which gives us the solutions:
  • \( x = 4 \)
  • \( x = -1 \)
These values of \( x \) satisfy the original equation, showing how effective the quadratic formula is in finding the roots of a quadratic equation.

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Most popular questions from this chapter

For the following exercises, describe how the graph of the function is a transformation of the graph of the original function \(f\) . $$ y=f(x-2)+3 $$

For the following exercises, determine the interval(s) on which the function is increasing and decreasing. $$ f(x)=4(x+1)^{2}-5 $$

For the following exercises, find functions \(f(x)\) and \(g(x)\) so the given function can be expressed as \(h(x)=f(g(x))\) $$ h(x)=\frac{3}{x-5} $$

For the following exercises, use the function values for \(f\) and \(g\) shown in Table 1.24 to evaluate the expressions. $$ \begin{array}{|c|c|c|c|c|c|c|c|}\hline x & {-3} & {-2} & {-1} & {0} & {1} & {2} & {3} \\ \hline f(x) & {11} & {9} & {7} & {5} & {3} & {1} & {-1} \\\ \hline g(x) & {-8} & {-3} & {0} & {1} & {0} & {-3} & {-8} \\\ \hline\end{array} $$ $$ (f \circ f)(3) $$

For the following exercises, find functions \(f(x)\) and \(g(x)\) so the given function can be expressed as \(h(x)=f(g(x))\) $$ h(x)=\frac{4}{(x+2)^{2}} $$
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