Problem 96 - Precalculus 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 two values of \(x\) that satisfy \(f(g(x)) = -5\) are \(x = 1\) and \(x = -4/3\).

Step 1: Understand the Composition of Functions

Given the functions \(f(x)=1-2x\) and \(g(x)=3x^{2}+x-1\), the composition of the functions \(f(g(x))\) is obtained by taking \(g(x)\) as the input to function \(f\), i.e., \(1 - 2g(x) = 1-2(3x^{2}+x-1)\).

Step 2: Simplify the Composite Function

Simplify the expression \(1-2(3x^{2}+x-1)\) by using distributive property, results in \(1 - 6x^{2} - 2x + 2 = -6x^{2} - 2x + 3\)

Step 3: Solve for x

Set \(f(g(x)) = -5\) by substituting \(-6x^{2} - 2x + 3\) to -5 i.e. \(-6x^{2} - 2x + 3 = -5\). The given equation simplifies to \(6x^{2} + 2x - 8 = 0\). Now apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\) where \(a=6\), \(b=2\), and \(c=-8\). \nBy doing so, \(x = \frac{-2 \pm \sqrt{(2)^{2}-4(6)(-8)}}{2(6)} = \frac{-2 \pm \sqrt{4+192}}{12} = \frac{-2 \pm \sqrt{196}}{12} = \frac{-2 \pm 14}{12}\). This gives two solutions, \(x = 1\) and \(x = -4/3\).

Step 4: Verify the solution

To ensure that \(x = 1\) and \(x = -4/3\) are the correct solutions, substitute each value of x back into the original equation \((f \circ g)(x)=-5\). If the equation holds, then the solutions are correct.

Key Concepts

Quadratic Formula

The quadratic formula is a fundamental tool used for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). This formula provides a way to find values of \( x \) that satisfy the quadratic equation, and it is expressed as:

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

Here, \( a \), \( b \), and \( c \) are coefficients of the equation. The plus-minus symbol (\( \pm \)) indicates that there generally will be two solutions to a quadratic equation.
The expression under the square root, \( b^2 - 4ac \), is known as the discriminant. The discriminant tells us about the nature of the roots:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is one real solution (roots are repeated).
• If \( b^2 - 4ac < 0 \), the solutions are complex and no real numbers will satisfy the equation.

In the exercise, the quadratic equation \( 6x^2 + 2x - 8 = 0 \) was solved using the quadratic formula. Calculating the discriminant, \( 4 + 192 = 196 \), which is positive, demonstrates that there are two real solutions: \( x = 1 \) and \( x = -\frac{4}{3} \). This step confirms that the solutions belong to the real number set.

Distributive Property

The distributive property is an essential algebraic property used to simplify expressions and solve equations. It is often implemented when multiplying a number by a sum in parentheses. In essence, the property dictates:

• \( a(b + c) = ab + ac \)

This means that you distribute the multiplication of \( a \) across each term within the parentheses. In the exercise, the expression \( 1 - 2(3x^2 + x - 1) \) requires the distributive property to simplify it.
Applying the distributive property, you multiply \(-2\) with each term inside the parentheses:

• \(-2 \times 3x^2 = -6x^2 \)
• \(-2 \times x = -2x \)
• \(-2 \times -1 = 2 \)

Combining these results gives the simplified expression \(-6x^2 - 2x + 2\). This step simplifies the problem and prepares the quadratic equation for the next stages of solving.

Function Verification

Function verification is a crucial step in ensuring that solutions obtained from solving equations are indeed correct. This process involves substituting the solutions back into the original function or equation to check for correctness.
In our exercise, after solving for \( x \) using the quadratic formula, we obtained \( x = 1 \) and \( x = -\frac{4}{3} \). To verify these solutions, each value of \( x \) was substituted back into the composition \( (f \circ g)(x) = -5 \).

• Substitute \( x = 1 \) back, compute \((f \circ g)(1)\) to see if it equals \(-5\).
• Substitute \( x = -\frac{4}{3} \), compute \((f \circ g)(-\frac{4}{3})\) and verify the same.

If both substitutions satisfy the original condition \((f \circ g)(x) = -5\), the solutions are confirmed correct. This step not only proves accuracy but also strengthens understanding of how solutions apply practically within the problem context.