Question

(a) which of the following functions have 5 in their domain?

$f\left(x\right)=x^2-3x,g\left(x\right)=\frac{x-5}{x},h\left(x\right)=\sqrt{x-10}$

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

Short answer

(a) The value of 5 is domain of f and g functions.

(b) The value is$f\left(5\right)=10,g\left(5\right)=0$

Step-by-step solution

Part a. Step 1. Use the definition of functions.

Function, is an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable).

A function is also defined as, given a set of inputs X and a set of possible outputs Y as a set of ordered pairs (x,y). We can write the statement that f is a function from X to Y using the function notatiof: $X\to Y$.

Part a. Step 2. find the function domains.

We are given that $f\left(x\right)=x^2-3xg\left(x\right)=\frac{x-5}{x}h\left(x\right)=\sqrt{x-10}$,

Since f is algebraic its domain is $\left(-\infty ,\infty \right)$and domain contains element $x=5$

Since $g\left(x\right)=\frac{x-5}{x}$and since denominator cannot be zero, its domain is $\left(-\infty ,0\right)\cup \left(0,\infty \right)$and domain contains element $x=5$

Since $h\left(x\right)=\sqrt{x-10}$and since radical cannot be negative we have

$\begin{array}{l}x-10\ge 0\\ x\ge 10\end{array}$,

Part a. Step 3. Evaluate the results.

The function domain is $\left(10,\infty \right)$ and domain does not contains element$x=5$

Part b. Step 1. Use the definition of functions.

Function, is an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable).

A function is also defined as, given a set of inputs X and a set of possible outputs Y as a set of ordered pairs (x,y). We can write the statement that f is a function from X to Y using the function notatiof: $X\to Y$.

Part b. Step 2. Analyze the information given.

We are given that $f\left(x\right)=x^2-3x$, to find we value at$x=5$ we plug the value of x in the function.

Part b. Step 3. Evaluate the results for function f.

We plug the value of x in the function as shown:

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

Part b. Step 3. Evaluate the results for function g.

We are given that $g\left(x\right)=\frac{x-5}{x}$, to find we value at $x=5$we plug the value of x in the function as shown:

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