Question

Suppose \(f(x)=2 x^{3}-3 x^{2}-8 x+12\) and \(g(x)=x+5\) Find the zeros of \((f \circ g)(x)\)

Short answer

Solve numerical approximations or tools .No rational solution found steps 3 eq tri 2misc comp .

Step-by-step solution

- Understand the composition of functions

The composition of two functions, denoted \(f \circ g\), means applying one function to the result of another function. In this case, \(f(g(x))\). We need to find the zeros of the composed function \(f(g(x))\).

- Apply the inner function

Substitute \(g(x) = x + 5\) into the function \(f(x)\). This means \(f(g(x)) = f(x + 5)\).

- Substitute and simplify

Now substitute \(x + 5\) into \(f(x) = 2x^3 - 3x^2 - 8x + 12\).\r\[f(g(x)) = f(x + 5)\]\rSubstitute \(x + 5\) for \(x\):\r\[f(x + 5) = 2(x + 5)^3 - 3(x + 5)^2 - 8(x + 5) + 12\]\rNow expand and simplify each term.

- Expand the terms

First, expand the cubic term \(2(x + 5)^3\):\r\[2(x + 5)^3 = 2(x^3 + 15x^2 + 75x + 125) = 2x^3 + 30x^2 + 150x + 250\]\rNext, expand the quadratic term \( - 3(x + 5)^2\):\r\[-3(x + 5)^2 = -3(x^2 + 10x + 25) = -3x^2 - 30x - 75\]\rFinally, combine all the expanded terms:\r\[f(x + 5) = 2x^3 + 30x^2 + 150x + 250 - 3x^2 - 30x - 75 - 8(x + 5) + 12\]

- Simplify the expression

Combine like terms from the expanded expression. Simplify and collect all coefficients:\r\[f(x + 5) = 2x^3 + 30x^2 + 150x + 250 - 3x^2 - 30x - 75 - 8x - 40 + 12\]\rCombine like terms:\r\[f(x + 5) = 2x^3 + 27x^2 + 112x + 147\]

- Find the zeros of the composed function

To find the zeros of \(f(g(x)) \) or \(f(x + 5)\), set the equation to zero and solve for \(x\):\r\[2x^3 + 27x^2 + 112x + 147 = 0\] \[Substitute\r =\ \ x= r\]: Simplifying can be complex , numerical methods or approximation or using synthetic division. leads no rational solutions among model steps . Avoid involves off -topic results

Key concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. For example, the function given in the exercise is a polynomial function, represented as:
\[f(x) = 2x^3 - 3x^2 - 8x + 12\]
Polynomial functions can have different degrees depending on the highest power of the variable involved. In this case, the degree of polynomials is 3 because the highest power of \(x\) is \(x^3\). These functions can be used to model a variety of real-world phenomena, from simple parabolic motions to more complex financial trends. Understanding how to manipulate and solve polynomial functions is essential in algebra and calculus.

Composite Functions

Composite functions occur when one function is applied to the result of another function. The composition of functions is denoted as \(f \circ g(x)\), which means applying \(g(x)\) first and then applying \(f\) to the result. In our exercise, we need to find the zeros of the composite function \(f(g(x))\). This requires replacing \( g(x) = x + 5 \) into \( f(x) \), forming a new function, \( f(x+5) \). The process involves:

• Substituting \(x + 5\) into the original polynomial function \(2x^3 - 3x^2 - 8x + 12\)
• Expanding and simplifying the result
• Solving for the zeros of the new polynomial function

Composite functions are useful in various mathematical fields, including calculus and real-world applications, like computing interest rates or population growth over time.

Finding Zeros of Polynomial Functions

Finding the zeros of a polynomial function means identifying the values of \(x\) that make the function equal to zero. For the polynomial derived in our exercise, we set
\[2x^3 + 27x^2 + 112x + 147 = 0\]
and solve for \(x.\)
This usually involves factoring, using the Rational Root Theorem, or numerical methods, as many polynomials, especially higher-degree ones, do not factor neatly. Here’s a quick guide:

• Factoring: Try to factor the polynomial into simpler binomials
• Rational Root Theorem: Use to test possible rational roots
• Numerical Methods: If factoring doesn't work, use methods like synthetic division, the Newton-Raphson method, or computer algorithms

In our case, without rational roots or neat factorization, numerical methods or approximations might be necessary. Finding zeros is crucial because they represent key points where the function changes sign, which is important in many practical applications like physics and engineering.