Answer to Question #179466 in Algebra for Paul Carpenter

Question

For this written assignment, answer the following questions showing all of your work.

1. Find the domain of the function using interval notation.

2. Sketch a graph of a piecewise function. Write the domain in interval notation.

[Suggestion: for example, go to www.desmos.com/calculator and write

for {-1 ≤ x ≤ 1}

and

y = 3x - 2 {1 ≤ x ≤ 3}

Then choose your own functions 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 determined 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

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The full question values

For (1)

$\boxed{f(x)=\dfrac{\sqrt{x}-6}{\sqrt{x}-4} }$ 

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(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 $\boxed{[0,\:16)\cup \left(16,\:\infty \:\right)}$

The graph

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(2) The graph of piece wise function 

$\boxed{f(x) = \begin{cases} x^2 & , \ -1<x<1 \\\\ 3x-2 & , \ 1<x<3\\ \end{cases}}$ 

Is shown for the given interval

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 $\boxed{(−1,1)∪(1,3)}$

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(3)

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

$$\boxed{c(0) = 10(0) + 500}$$ $$\boxed{c(0)=500}$$

So, the fixed cost is $\boxed{\$500}$

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(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 $\boxed{\$750}$

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(c)

Since the maximum cost allowed is $\$1500$  

$$\boxed{10x+500 \leq 1500}$$

To solve this inequality

First, subtract 500 from both sides