Problem 37 - College Algebra with Corequisite Support

Question

For the following exercises, evaluate the function \(f\) at the indicated values \(f(-3), f(2), f(-a),-f(a), f(a+h)\). Given the function \(f(x)=x^{2}-3 x:\) (a) Evaluate \(f(5)\). (b) Solve \(f(x)=4\).

Step-by-Step Solution

Answer

(a) 10. (b) Solutions: 4 and -1.

Step 1: Understand the Function Definition

The function given is \( f(x) = x^2 - 3x \). We will use this definition to evaluate the function at different specified values, and to solve for when \( f(x) = 4 \).

Step 2: Evaluate f(5)

To find \( f(5) \), substitute \( x = 5 \) into the function. \[ f(5) = 5^2 - 3(5) \]\[ f(5) = 25 - 15 \]\[ f(5) = 10 \]So, the function evaluated at \( x = 5 \) is 10.

Step 3: Solve f(x) = 4

To solve \( f(x) = 4 \), set the function equal to 4 and solve for \( x \):\[ x^2 - 3x = 4 \]Bring 4 to the left side to form a quadratic equation:\[ x^2 - 3x - 4 = 0 \]Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -3, c = -4 \):\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]\[ x = \frac{3 \pm \sqrt{9 + 16}}{2} \]\[ x = \frac{3 \pm \sqrt{25}}{2} \]\[ x = \frac{3 \pm 5}{2} \]Thus, the solutions are:\[ x = \frac{8}{2} = 4 \]\[ x = \frac{-2}{2} = -1 \]

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to solve any quadratic equation, which takes the form \( ax^2 + bx + c = 0 \). Its formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Let's break this down:- \( a \) is the coefficient of \( x^2 \).- \( b \) is the coefficient of \( x \).- \( c \) is the constant term.- The term \( \sqrt{b^2 - 4ac} \) is known as the discriminant.The discriminant determines the nature of solutions:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is one real solution (a repeated root).
• If \( b^2 - 4ac < 0 \), there are no real solutions, but two complex solutions.

Using the quadratic formula, you can solve quadratic equations quickly, even when factoring is challenging.

Solving Quadratic Equations

Solving quadratic equations, like the provided equation \( x^2 - 3x = 4 \), is often necessary in mathematics. It involves finding the values of \( x \) that satisfy the equation.In our problem, we first rearrange the terms to get:\[x^2 - 3x - 4 = 0\]This transforms the equation into the standard quadratic form. We can now apply the quadratic formula.Substituting into the formula:

• \( a = 1, b = -3, c = -4 \)
• The discriminant becomes \( 9 + 16 = 25 \)

Calculating further, \( \sqrt{25} = 5 \), provides two solutions:- \( x = \frac{3 + 5}{2} = 4 \)- \( x = \frac{3 - 5}{2} = -1 \)Hence, the solutions \( x = 4 \) and \( x = -1 \) make the equation true when substituted back into the original quadratic equation.

Substitution Method

The substitution method is particularly useful for evaluating functions at specific points. It involves replacing the variable \( x \) with a given value.Here’s how it works with an example:Given the function \( f(x) = x^2 - 3x \), to evaluate \( f(5) \), substitute 5 for \( x \):\[f(5) = 5^2 - 3(5)\]Calculating step by step:

• \( 5^2 = 25 \)
• \( 3 \times 5 = 15 \)
• \( 25 - 15 = 10 \)

So, \( f(5) = 10 \). This substitution process applies to evaluating any function \( f \) at a point, ensuring we verify the function's behavior at specific values.