Chapter 2.1, Problem 3E

1. (a) Which of the following functions have 5 in their domain?

   $\begin{array}{ccc}f(x)=x^2-3x& g(x)=\frac{x-5}{x}& h(x)=\sqrt{x-10}\end{array}$

2. (b) For the functions from part (a) that do have 5 in their domain, find the value of the function at 5.

To determine

The functions which have 5 in their domain.

Answer to Problem 3E

The functions which have 5 in their domain are $f\left(x\right)=x^2-3x$ and $g\left(x\right)=\frac{x-5}{x}$.

Explanation of Solution

Given:

The given functions are,

$f\left(x\right)=x^2-3x$ (1)

$g\left(x\right)=\frac{x-5}{x}$   (2)

$h\left(x\right)=\sqrt{x-10}$ (3)

Calculation:

The set of possible inputs for a function f is called the Domain of f.

For any real number the function $f\left(x\right)=x^2-3x$ gives a real value.

Therefore, the set of all real number $ℝ$ is the domain of $f\left(x\right)=x^2-3x$.

Since 5 is a real number, therefore 5 is the domain of $f\left(x\right)=x^2-3x$.

For any real number except zero the function $g\left(x\right)=\frac{x-5}{x}$ gives a real value.

Therefore, the set of all real number except zero $ℝ-\left\{0\right\}$ is the domain of $g\left(x\right)=\frac{x-5}{x}$.

Since 5 is a real number except zero, therefore 5 is the domain of $g\left(x\right)=\frac{x-5}{x}$.

For any real number greater than and equal to 10 the function $h\left(x\right)=\sqrt{x-10}$ gives a real value.

Therefore, the set of all real number greater than and equal to 10 is the domain of $h\left(x\right)=\sqrt{x-10}$.

Since 5 is not a real number greater than and equal to 10, therefore 5 is not the domain of $h\left(x\right)=\sqrt{x-10}$.

Thus, the functions which have 5 in their domain are $f\left(x\right)=x^2-3x$ and $g\left(x\right)=\frac{x-5}{x}$.

To determine

The values of functions at $x=5$ who have 5 in their domain.

Answer to Problem 3E

The values of $f\left(x\right)=x^2-3x$ at $x=5$ is 10 and the value of $g\left(x\right)=\frac{x-5}{x}$ at $x=5$ is 0.

Explanation of Solution

Given:

From part (a), the functions which have 5 in their domain are given below,

$f\left(x\right)=x^2-3x$

$g\left(x\right)=\frac{x-5}{x}$

Calculation:

Substitute 5 for x in equation (1), to find the value of $f\left(x\right)=x^2-3x$ at $x=5$,

$\begin{array}{c}f\left(5\right)={\left(5\right)}^2-3\left(5\right)\\ =25-15\\ =10\end{array}$

For $x=5$ the function $f\left(x\right)=x^2-3x$ has a value 10.

Substitute 5 for x in equation (2) to find the value of $g\left(x\right)=\frac{x-5}{x}$ at $x=5$,

$\begin{array}{c}g\left(5\right)=\frac{\left(5\right)-5}{\left(5\right)}\\ =\frac{0}{5}\\ =0\end{array}$

For $x=5$ the function $g\left(x\right)=\frac{x-5}{x}$ has a real value 0.

Thus, the values of $f\left(x\right)=x^2-3x$ at $x=5$ is 10 and the value of $g\left(x\right)=\frac{x-5}{x}$ at $x=5$ is 0.