Answer to Question #123076 in Algebra for Dani wagas

Question

1. Find the domain of the function using interval notation. f(x) = √x - 6/√x-4
2. Sketch a graph of a piecewise function. Write the domain in interval notation. [go to www.desmos.com/calculator and write y = x^2 {-1<x<1} and y = 3x - 2 {1<x<3}. Then choose your own function and have fun.]
3. The cost in dollars of making x items is given by the function c(x) = 10x + 500.
a. The fixed cost is determine when zero items are produced. Find the fixed cost for this item.
b. What is the cost of making 25 items?
c. Suppose the maximum cost allowed is $1500. What are the domain and range of the cost function, c(x)?

Answer 1

(1) To find the he domain of the function $f(x)=\dfrac{\sqrt{x}-6}{\sqrt{x}-4}$ graph the function.

The domain of a graph consists of all the input values shown on the x-axis.

Domain in interval notation is $[0,\:16)\cup \left(16,\:\infty \:\right)$

See the graph below

(2) The graph of piece wise function $f(x) = \begin{cases} x^2 & , \ -1<x<1 \\\\ 3x-2 & , \ 1<x<3\\ \end{cases}$ is shown below

The domain of a graph consists of all the input values shown on the x-axis.

The function undefined at -1 , 1 and 3

So, the domain in interval notation is $(-1,1) \cup (1,3)$

(3)

(a) To find the fixed cost substitute $x=0$ in the equation $c(x) = 10x + 500$

$$c(0) = 10(0) + 500$$

$$c(0)=500$$

So, the fixed cost is $\$500$

(b) To find the cost of making 25 items substitute $x=25$ in the equation

$$c(x) = 10x + 500\\\\ c(25) = 10(25) + 500\\\\ c(25)= 750$$

the cost of making 25 items is $\$750$

(c)

Since the maximum cost allowed is $\$1500$  

$$10x+500 \leq 1500$$

To solve this inequality

First, subtract 500 from both sides

$$10x\leq 1000$$

Divide both sides by 10

$$x\leq100$$

This means you can make at most 100 items.

The domain is $[0,100].$

The range is $[0,1500]$ .