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A function f has domain [0, 2] and range [0, 1]. (The notation [a, b] denotes {x : a ≤ x ≤ b}.) Define the function g as g(x) = 1 − f(x + 1). If [p, q] is the domain of g and [r,s] is the range of g, then determine p + q + r + s.

 Oct 31, 2020

Best Answer 

 #2
avatar+111600 
+2

A function f has domain [0, 2] and range [0, 1]. (The notation [a, b] denotes {x : a ≤ x ≤ b}.)

Define the function g as g(x) = 1 − f(x + 1).

If [p, q] is the domain of g and [r,s] is the range of g, then determine p + q + r + s.

 

I am not really sure but,

 

f(x) has a domain of  [0,2]

so

\(For\;\;f(x)\quad0\le x\le2\\ For\;\;f(x+1)\\ 0\le x+1 \le2\\ -1\le x \le1\)

 

So the domain of g is [-1,1]      p=-1,  q=1

----

 

\(0\le f(x+1) \le 1\\ \text{seperates to}\\ 0\le f(x+1) \qquad \qquad f(x+1) \le 1\\ 0\ge -f(x+1) \quad \qquad -f(x+1) \ge -1\\ 1\ge 1-f(x+1) \qquad 1-f(x+1) \ge 0\\ 1-f(x+1)\le 1 \qquad 0\le1-f(x+1)\\ 0\le1-f(x+1)\le 1\\ 0\le g \le 1\qquad \qquad \\\text{The range of g is [0,1]} \)

 

-1+1+0+1= 1

 

----------------------

 

Here is a graph that fits the description to help you understand what I have done

I chose f to be the top half of a circle, radius 1 and centre (1,0)

 

Here is the direct link to the Desmos graph if you want it.

https://www.desmos.com/calculator/xoul34kiqs

 

 

 

LaTex:

 

For\;\;f(x)\quad0\le x\le2\\
For\;\;f(x+1)\\
0\le x+1 \le2\\
-1\le x \le1

 

0\le f(x+1) \le 1\\
\text{seperates to}\\
0\le f(x+1) \qquad \qquad f(x+1) \le 1\\
0\ge -f(x+1) \quad \qquad -f(x+1) \ge -1\\
1\ge 1-f(x+1) \qquad 1-f(x+1) \ge 0\\
1-f(x+1)\le 1 \qquad  0\le1-f(x+1)\\
0\le1-f(x+1)\le 1\\
0\le g \le 1\qquad \qquad \\\text{The range of g is [0,1]}

 Nov 1, 2020
 #2
avatar+111600 
+2
Best Answer

A function f has domain [0, 2] and range [0, 1]. (The notation [a, b] denotes {x : a ≤ x ≤ b}.)

Define the function g as g(x) = 1 − f(x + 1).

If [p, q] is the domain of g and [r,s] is the range of g, then determine p + q + r + s.

 

I am not really sure but,

 

f(x) has a domain of  [0,2]

so

\(For\;\;f(x)\quad0\le x\le2\\ For\;\;f(x+1)\\ 0\le x+1 \le2\\ -1\le x \le1\)

 

So the domain of g is [-1,1]      p=-1,  q=1

----

 

\(0\le f(x+1) \le 1\\ \text{seperates to}\\ 0\le f(x+1) \qquad \qquad f(x+1) \le 1\\ 0\ge -f(x+1) \quad \qquad -f(x+1) \ge -1\\ 1\ge 1-f(x+1) \qquad 1-f(x+1) \ge 0\\ 1-f(x+1)\le 1 \qquad 0\le1-f(x+1)\\ 0\le1-f(x+1)\le 1\\ 0\le g \le 1\qquad \qquad \\\text{The range of g is [0,1]} \)

 

-1+1+0+1= 1

 

----------------------

 

Here is a graph that fits the description to help you understand what I have done

I chose f to be the top half of a circle, radius 1 and centre (1,0)

 

Here is the direct link to the Desmos graph if you want it.

https://www.desmos.com/calculator/xoul34kiqs

 

 

 

LaTex:

 

For\;\;f(x)\quad0\le x\le2\\
For\;\;f(x+1)\\
0\le x+1 \le2\\
-1\le x \le1

 

0\le f(x+1) \le 1\\
\text{seperates to}\\
0\le f(x+1) \qquad \qquad f(x+1) \le 1\\
0\ge -f(x+1) \quad \qquad -f(x+1) \ge -1\\
1\ge 1-f(x+1) \qquad 1-f(x+1) \ge 0\\
1-f(x+1)\le 1 \qquad  0\le1-f(x+1)\\
0\le1-f(x+1)\le 1\\
0\le g \le 1\qquad \qquad \\\text{The range of g is [0,1]}

Melody Nov 1, 2020
 #3
avatar+61 
+1

Thanks.

AaronWang2004  Nov 1, 2020
 #4
avatar+111600 
0

Do you understand it?

Do you have any questions about what I have written?

Will you be able to do these by yourself in future?

 

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Melody  Nov 1, 2020

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