**High School Number Sense Lesson 78: Range of a Function**

We will revisit the world of Calculus today, but don't worry--it's only temporary. We have already looked at several different concepts surrounding

**functions**and their properties. Today we will look at how to determine their maximum and/or minimum values, as well as their ranges. This concept appeared

**7 times**last year, with a median placement at

**question # 77**.

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The

**range**of a function consists of the possible set of outputs (or y values) based on the given inputs. The

**maximum value**of a function is the largest possible value of y, and the

**minimum value**of a function is the smallest possible value of y.

**How to Solve:**

*To find the maximum or minimum value of a function:*

**ax**

^{2}+ bx + c = 02. Use this equation to solve for the min or max value:

**(4ac - b**

^{2})/4a3. If a > 0, you have found the minimum. If a < 0, you have found the maximum.

**Example 1: The range of the function y = |2x| - 3 is y ≥ ___**

- By definition, we know that the absolute value of anything has to be greater than 0.
- Replace |2x| with 0, and we are left with
**y = 0 - 3**, which is**-3**. This is our answer.

**Example 2:**

**The range of the function y = √(3 - x) is y ≥ ___**

1. We know that the square root of any number cannot be negative.

2. Since everything in the function is under the radicand (square root sign), the answer has to be

**0**.

**Example 3:**

**The range of the function y = -x**

^{4}+ 4 is y ≤ ___1. We know that the bigger x gets, the bigger x

^{4}gets, and therefore the smaller

**-x**gets. (There is no limit to how small).

^{4}2. So let's look at the other end of our range... The smallest that x

^{4}can get is 0. We are adding 4 in the function, so the answer must be

**4**.

**Example 4:**

**The range of the function y = x**

^{2}- x - 2 is y ≥ ___1. In the format

**y = ax**, a = 1, b = -1, and c = -2.

^{2}+ bx + c2. Let's solve the numerator first:

**4ac - b**= 4(1)(-2) - (-1)

^{2}^{2}= -8 - 1 =

**-9**.

3. Now for the denominator:

**4a**= 4(1) =

**4**.

4. The answer is

**-9/4**, which does not reduce.

**Example 5:**

**The minimum value of f(x) = 5(x - 3)**

^{2}+ 2 is ___1. First, square (x - 3) to get

**x**. Multiply this by 5 and add 2 to get

^{2}- 6x + 9**5x**=

^{2}- 30x + 45 + 2**5x**.

^{2}- 30x + 472. In the format

**y = ax**, a = 5, b = -30, and c = 47.

^{2}+ bx + c3. Let's solve the numerator:

**4ac - b**= 4(5)(47) - (-30)

^{2}^{2}= 20(47) - 900 = 940 - 900 =

**40**.

4. Now the denominator:

**4a**= 4(5) =

**20**.

5. The answer is

**40/20**, which reduces to

**2**.

*Notice that we could have used logic on this one as well...the smallest value for anything squared is 0, so the smallest value for*

**5(x - 3)**= 0. Add 2 to get^{2}**2**, which was our answer.**Check back soon for a free worksheet to help you practice FunctionRange.**

**Up Next for High School: EstMult142857**