Question (math - inequalities)

In summary, when finding the range of a function in a specific domain, the approach is to find the lowest and highest possible values of the function in that domain. In the first case, the range is { y | 4 <= y <= 22 } because the lowest value of y (4) is found when x=3, and the highest value of y (22) is found when x=0 or x=6. In the second case, the range is {y | 3 <= y <= 12} because the lowest value of y (3) is found when x=-3, and the highest value of y (12) is found when x=3. It is important to carefully consider the domain and the
  • #1
preet
98
0
What do you in these cases:

A function is g(x) = 2 (x-3) ^2 + 4
Find the range when 0 <= x <= 6

steps i took:
0 <= x <=6
0-3 <= x-3 <= 6-3
-3^2 <= (x-3)^2 <= 3^2
9 <= (x-3)^2 <= 9
22 <= 2 (x-3)^2 + 4 <=22
:: what did i do wrong?

correct answer is:
{ y | 4 <= y <= 22 }

---------------------
case#2

A certain function is defined as f(x)=x^2 + 3
if -3 <= x <=3, determine the range

i got {y | 3 <= y <= 9 }
correct answer is:
{y | 3 <= y <= 12}

Thanks in advance
Preet
 
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  • #2
The range for a certain function in a certain domain will be the lowest value of y to the highest value. Let's start with case 2.

You have the lower limit right, the lowest that function will ever be (in the domain) is 3. Now the upper limit, can the function not be 3^2 + 3 = 12?

For the first case, you are not just looking at the value of y that corresponds to the upper and lower limits of the domain, you are looking for the LOWEST and the HIGHEST y values in that domain. The upper limit you got right. Now the lower limit, is there any number between 0 and 6 that you can use to make you answer lower than 22?
 
  • #3
preet said:
0-3 <= x-3
-3^2 <= (x-3)^2
:: what did i do wrong?

a>b does not imply [itex]a^2 > b^2 [/itex], unless both are positive.

Try some examples :
2 > -3, but 4 < 9
-2 > -3, but 4 < 9
 
  • #4
The second problem is pretty straightforward. I don't see how you could have got 9 instead of 12. Why don't you show the steps; like you did with the first problem. This will make it easier to diagnose the error.

And as for the first problem, the approach you used is not the way to solve it. Ask yourself what values of x in the domain maximize and minimize g(x).

Clearly g(x) is mimimal, for the smallest value that (x-3)^2 can take, which in turn is minimized by finding the smallest possible value that |x-3| can take. Clearly, this happens when x=3. At this value of x, what value does g(x) take ?
 
  • #5
f(x)=x^2 + 3

domain:-3 <= x <=3
the way shown in the book is to turn the domain into the range by performing all of the operations in the function (domain -> function -> range)
...
-3 <= x <= 3
-3^2 <= x^2 <= 3^2
9 + 3 <= x^2 <= 9+3
12 <= x^2 <= 12

I don't even remember what I'm doing... my exam is coming up and I forgot all of this stuff... my answer doesn't even make sense. I don't know how I got the previous answer.

Is there some algebraic method of doing this or do you have to think every time which value will give you the min/max values? Even looking at the first one, shouldn't the range be 3 <= y <= 22? 3 is between 6 and 0 and gives the lower y value... but the answer in the book says 4. If you want to apply the method where you apply all of the operations of the function, do you have to find like the greatest and least absolute value or something? TiA.
 
  • #6
preet said:
f(x)=x^2 + 3

domain:-3 <= x <=3
the way shown in the book is to turn the domain into the range by performing all of the operations in the function (domain -> function -> range)
...
-3 <= x <= 3
-3^2 <= x^2 <= 3^2
9 + 3 <= x^2 <= 9+3
12 <= x^2 <= 12

Please read my post. I've told you why you can't do this.

Is there some algebraic method of doing this or do you have to think every time which value will give you the min/max values? Even looking at the first one, shouldn't the range be 3 <= y <= 22? 3 is between 6 and 0 and gives the lower y value... but the answer in the book says 4.
Yes, y= g(x) = 4, when x=3. Please read my posts slowly and carefully. You're panicking, and that just doesn't help you one bit. Calm down and start over.
 

What are inequalities in math?

Inequalities in math are mathematical statements that compare two quantities or expressions. They use symbols such as <, >, ≤, and ≥ to show that one quantity is less than, greater than, less than or equal to, or greater than or equal to another quantity.

How do you solve inequalities?

To solve inequalities, you follow the same rules as solving equations. You can add, subtract, multiply, or divide both sides of the inequality by the same number, as long as the inequality sign remains the same. However, if you multiply or divide by a negative number, you must flip the direction of the inequality sign.

What is the difference between solving an equation and solving an inequality?

The main difference between solving an equation and solving an inequality is that when solving an inequality, you have to pay attention to the direction of the inequality sign. Also, when multiplying or dividing by a negative number, you have to flip the direction of the sign.

Why are inequalities important in math?

Inequalities are important in math because they allow us to compare quantities and make statements about their relationships. They are used in many real-world scenarios, such as measuring inequalities in income, population, and temperature, among others.

What are some common mistakes students make when solving inequalities?

Some common mistakes students make when solving inequalities include forgetting to flip the direction of the inequality sign when multiplying or dividing by a negative number, not following the correct order of operations, and not checking their solution in the original inequality. It is also important to remember that the solution to an inequality can be a range of values, not just a single number.

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