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Magazine Chapter 5: Problem 6
Use the given pair of functions to find the following values if they exist. $$\bullet (g \circ f)(0)$$ $$\bullet (f \circ g)(-1)$$ $$\bullet (f \circ f)(2)$$ $$\bullet (g \circ f)(-3)$$ $$\bullet (f \circ g)\left(\frac{1}{2}\right)$$ $$\bullet (f \circ f)(-2)$$ $$f(x)=\sqrt{3-x}, g(x)=x^{2}+1$$
Short Answer
Expert verified
1. 4
2. 1
3. \(\sqrt{2}\)
4. 7
5. \(\frac{\sqrt{7}}{2}\)
6. Undefined
Step by step solution
01
Understand the Notation
The notation \((g \circ f)(x)\) and \((f \circ g)(x)\) represent function compositions. Specifically, \((g \circ f)(x) = g(f(x))\) means you first apply the function \(f\) to \(x\), then apply the function \(g\) to the result. Similarly, \((f \circ g)(x) = f(g(x))\) means applying \(g\) to \(x\) first, then \(f\) to that result.
02
Evaluate (g 鈭� f)(0)
Calculate \(f(0)\) using \(f(x) = \sqrt{3-x}\): \(f(0) = \sqrt{3-0} = \sqrt{3}\). Then, evaluate \(g(\sqrt{3})\) with \(g(x) = x^2+1\): \(g(\sqrt{3}) = (\sqrt{3})^2+1 = 3+1 = 4\). Therefore, \((g \circ f)(0) = 4\).
03
Evaluate (f 鈭� g)(-1)
Calculate \(g(-1)\) using \(g(x) = x^2+1\): \(g(-1) = (-1)^2+1 = 1+1 = 2\). Then, evaluate \(f(2)\) using \(f(x) = \sqrt{3-x}\): \(f(2) = \sqrt{3-2} = \sqrt{1} = 1\). Therefore, \((f \circ g)(-1) = 1\).
04
Evaluate (f 鈭� f)(2)
Calculate \(f(2)\) using \(f(x) = \sqrt{3-x}\): \(f(2) = \sqrt{3-2} = \sqrt{1} = 1\). Then, evaluate \(f(1)\) using the same function: \(f(1) = \sqrt{3-1} = \sqrt{2}\). Therefore, \((f \circ f)(2) = \sqrt{2}\).
05
Evaluate (g 鈭� f)(-3)
Since \(f(x) = \sqrt{3-x}\), \(f(-3) = \sqrt{3-(-3)} = \sqrt{6}\). Then, evaluate \(g(\sqrt{6})\) using \(g(x) = x^2+1\): \(g(\sqrt{6}) = (\sqrt{6})^2 + 1 = 6 + 1 = 7\). Therefore, \((g \circ f)(-3) = 7\).
06
Evaluate (f 鈭� g)(\frac{1}{2})
Calculate \(g\left(\frac{1}{2}\right)\) using \(g(x) = x^2+1\): \(g\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2+1 = \frac{1}{4}+1 = \frac{5}{4}\). Then, evaluate \(f\left(\frac{5}{4}\right)\) using \(f(x) = \sqrt{3-x}\): \(f\left(\frac{5}{4}\right) = \sqrt{3-\frac{5}{4}} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2}\). Therefore, \((f \circ g)\left(\frac{1}{2}\right) = \frac{\sqrt{7}}{2}\).
07
Evaluate (f 鈭� f)(-2)
Calculate \(f(-2)\) using \(f(x) = \sqrt{3-x}\): \(f(-2) = \sqrt{3-(-2)} = \sqrt{5}\). Then, evaluate \(f(\sqrt{5})\) which is \(\sqrt{3-\sqrt{5}}\). Since \(\sqrt{5} > 3\), \(3-\sqrt{5}\) is negative, so \(f(\sqrt{5})\) is undefined since the square root of a negative number is not real. Therefore, \((f \circ f)(-2)\) is undefined.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain and Range
Understanding the domain and range of a function helps you know where the function can be applied and what kind of output it will yield. The **domain** of a function is the set of all input values (or x-values) for which the function is defined. For example:
The **range** is the set of all possible outputs (or y-values). The range varies depending on the function's characteristics:
- The quadratic function, like \( g(x) = x^2 + 1 \), accepts all real numbers. So its domain is all real numbers (\(-\infty, \infty\)).
- On the other hand, the square root function, \( f(x) = \sqrt{3-x} \), is only defined when \(3-x \geq 0\). Thus, the domain of \( f(x) \) is \(x \leq 3\).
The **range** is the set of all possible outputs (or y-values). The range varies depending on the function's characteristics:
- For \( g(x) = x^2 + 1 \), since the square of any real number is non-negative, \( g(x) \) will never be less than 1. Therefore, the range is \([1, \infty)\).
- For \( f(x) = \sqrt{3-x} \), possible outputs start from zero upwards, thus the range is \([0, \infty)\).
Evaluating Functions
Evaluating functions means finding the output of a function given a specific input. To evaluate a function like \( f(x) = \sqrt{3-x} \) or \( g(x) = x^2 + 1 \), follow these simple steps:
For instance, to find \( f(0) \), plug 0 into the function \( f(x) \):
\( f(0) = \sqrt{3-0} = \sqrt{3} \).
Similarly, to evaluate \( g(-1) \), substitute \(-1\) into \( g(x) \):
\( g(-1) = (-1)^2 + 1 = 1 + 1 = 2 \).
These steps help to break down even the most complicated-looking expressions.
- Substitute the given input value into the function.
- Simplify the equation to get the output.
For instance, to find \( f(0) \), plug 0 into the function \( f(x) \):
\( f(0) = \sqrt{3-0} = \sqrt{3} \).
Similarly, to evaluate \( g(-1) \), substitute \(-1\) into \( g(x) \):
\( g(-1) = (-1)^2 + 1 = 1 + 1 = 2 \).
These steps help to break down even the most complicated-looking expressions.
Square Root Function
The square root function is unique and requires careful handling because it has restrictions on its domain. A basic square root function looks like \( f(x) = \sqrt{x} \). However, in the case of \( f(x) = \sqrt{3-x} \), it is slightly more complex:
For example, \( f(2) = \sqrt{3-2} = \sqrt{1} = 1 \), a valid real number. However, if you attempt to evaluate \( f(4) \), it becomes \( \sqrt{3-4} = \sqrt{-1} \), which is undefined in the set of real numbers.
- The expression under the square root, \(3-x\), must be non-negative, resulting in the domain \(x \leq 3\).
- When evaluating, always ensure that the value inside the square root remains non-negative, as square roots of negative numbers are not real numbers.
For example, \( f(2) = \sqrt{3-2} = \sqrt{1} = 1 \), a valid real number. However, if you attempt to evaluate \( f(4) \), it becomes \( \sqrt{3-4} = \sqrt{-1} \), which is undefined in the set of real numbers.
Quadratic Function
Quadratic functions are polynomial functions of degree 2, usually written in the form \( g(x) = ax^2 + bx + c \). In our original problem, \( g(x) = x^2 + 1 \), a quadratic function without the linear \( b \) term (or it just equals zero).
This vertex is crucial because it signifies the lowest value of the function, so the range is \([1, \infty)\). Quadratic functions are often used in various applications because of their ability to model situations that involve change at a constant rate.
- The graph of a quadratic function is a parabola. In this case, since \( a = 1 \), the parabola opens upwards.
- The vertex of this parabola is at the lowest point and, for this particular function, it's at \( (0, 1) \) since \( g(0) = 0^2 + 1 = 1 \).
This vertex is crucial because it signifies the lowest value of the function, so the range is \([1, \infty)\). Quadratic functions are often used in various applications because of their ability to model situations that involve change at a constant rate.
