Problem 96 - College Algebra Essentials

Question

Find all values of \(x\) satisfying the given conditions. $$f(x)=1-2 x, g(x)=3 x^{2}+x-1, \text { and }(f \circ g)(x)=-5.$$

Step-by-Step Solution

Answer

The values of x are \(x = 1\) and \(x = -\frac{4}{3}\).

Step 1: Compute the function \(f(g(x))\)

In order to solve the problem, you have to first compute the composition of function \(f(g(x))\), which means you have to plug \(g(x)\) into \(f(x)\). So, this becomes \(f(g(x)) = 1 - 2g(x)\). Since you know that \(g(x) = 3x^2 + x - 1\), you can substitute this into the equation, making it \(f(g(x)) = 1 - 2(3x^2 + x - 1)\). After simplifying, you will get \(f(g(x)) = 1 - 6x^2 - 2x + 2 = -6x^2 - 2x + 3\).

Step 2: Applying given condition

By using the condition that \(f(g(x)) = -5\), you substitute that into the equation that was obtained in step 1: \(-6x^2 - 2x + 3 = -5\). After rearranging this equation, you will get a quadratic equation: \(6x^2 + 2x - 8 = 0\). You can simplify it further by dividing the entire equation by 2 to get \(3x^2 + x - 4 = 0\).

Step 3: Solving the quadratic equation

To find the x-values, the quadratic equation \(3x^2 + x - 4 = 0\) needs to be solved. For this, you can use the general quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), here a=3, b=1, c=-4. Substituting these values in the quadratic formula, you get \(x = \frac{-1 \pm \sqrt{1^2 - 4*3*(-4)}}{2*3}\), which simplifies to \(x = \frac{-1 \pm \sqrt{49}}{6}\) so the solutions are \(x = 1\) and \(x = -\frac{4}{3}\).

Key Concepts

Solving Quadratic Equations

Understanding how to solve quadratic equations is a fundamental skill in algebra. A quadratic equation is any equation that can be written in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\). To solve these equations, one can factorize, complete the square, or use the quadratic formula.

The process involves identifying the coefficients \(a\), \(b\), and \(c\), and determining the \(x\) values that satisfy the equation. With practice, you'll recognize common patterns and learn which methods are most efficient based on the equation's structure. In the case where factoring is complex or not possible, the quadratic formula provides a reliable method for finding the roots of any quadratic equation.

Quadratic Formula

The quadratic formula is a powerful tool that allows you to find the solutions to any quadratic equation, even when factoring is not feasible. It is derived from the process of completing the square and is written as \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\).

The formula includes a 'plus or minus' symbol, which indicates that there are typically two possible solutions, also known as roots. The expression under the square root, \(b^2 - 4ac\), known as the discriminant, can give you insights into the nature of the roots; if positive, there are two real solutions, if zero, there's exactly one real solution, and if negative, the solutions are complex.

Algebraic Functions

In algebra, functions are used to represent relationships between two sets of numbers or variables. Algebraic functions are rules that assign each input exactly one output, and they can be expressed by equations involving polynomial expressions, radicals, and rational expressions.

Function composition, denoted as \(f \circ g\) and read as 'f composed with g', is a particular operation where you apply one function to the results of another function. In the provided exercise, \(f \circ g)(x)\) was given a specific value, which sets up an equation that can be solved for \(x\). Such problems often involve substituting one function into another and then solving the resulting algebraic equation, which in this case, is a quadratic equation.