Problem 10 - Algebra A Combined Function

Question

If \(f(x)=x^{2}-6 x+2, g(x)=-2 x\), and \(h(x)=\sqrt{x}\), find each composition. $$ (h \circ f)(-2) $$

Step-by-Step Solution

Answer

The composition \((h \circ f)(-2)\) is \(3\sqrt{2}\).

Step 1: Understand the composition

The problem asks us to find the composition \((h \circ f)(-2)\), which means we need to first evaluate \(f(-2)\) and then apply the function \(h(x)\) to that result. This is expressed as \(h(f(-2))\).

Step 2: Evaluate \(f(-2)\)

Plug \(x = -2\) into the function \(f(x) = x^2 - 6x + 2\). This gives us:\[f(-2) = (-2)^2 - 6(-2) + 2 = 4 + 12 + 2 = 18\]

Step 3: Apply \(h(x)\) to \(f(-2)\)

Now, take the result from Step 2, which is \(18\), and substitute it into \(h(x) = \sqrt{x}\). This gives us:\[h(18) = \sqrt{18} = 3\sqrt{2}\]

Step 4: Simplify the result

The expression \(3\sqrt{2}\) is already in its simplest form, so there is no further simplification needed.

Key Concepts

Polynomial Functions

Polynomial functions are a fundamental concept in algebra and mathematics as a whole. A polynomial function is essentially an expression consisting of variables, coefficients, and exponents.
These exponents must be whole numbers, meaning they cannot be fractions or negative numbers.

• For example, a simple polynomial function is given by: \( f(x) = x^2 - 6x + 2 \).
• Here, the highest exponent (also called the degree) is 2, making it a quadratic polynomial function.

Polynomial functions can range from simple linear equations to complex higher-degree polynomials. The degree of the polynomial determines the shape of its graph:

• Linear polynomials (degree 1) form straight lines.
• Quadratic polynomials (degree 2) form parabolas.

Understanding how to work with polynomial functions involves evaluating them at various points, as seen in the original exercise where \( f(-2) \) was calculated.

Square Root Function

The square root function is another essential component of algebra, recognizable by its radical symbol \( \sqrt{} \).
It primarily deals with finding a number that, when multiplied by itself, gives the original number inside the radical.

• In the exercise provided, the function is represented as \( h(x) = \sqrt{x} \).
• This highlights the need to extract the principal square root of a given number.

Mathematically, the square root function is not defined for negative numbers within the real number system because there's no real number that can be squared to produce a negative number. However, it is readily applicable when dealing with positive results such as when \( \sqrt{18} \) was evaluated.
The process involves simplifying the expression under the square root when possible, such as expressing \( \sqrt{18} \) as \( 3\sqrt{2} \), which is its simplest radical form.

Evaluation of Functions

Evaluation of functions is a process that involves substituting a determined value into a function and determining its resulting output.
This step is crucial in analyzing how functions behave under various transformations and conditions.
When evaluating a function, you essentially replace the variable with a specific number:

• For example, evaluating \( f(-2) \) involves plugging \(-2\) into the polynomial \( f(x) = x^2 - 6x + 2 \), which results in \( 18 \).
• Similarly, in the exercise, this result was evaluated further using the square root function \( h(x) \), resulting in \( \sqrt{18} \).

The evaluation process helps in tasks such as finding function compositions.
As seen in the original exercise, understanding the composition \((h \circ f)(x)\) required evaluating \(h(f(x))\).
This practice is vital in solving advanced problems and functions as it allows for deeper insights into how changes in inputs affect outputs.