Math  /  Algebra

QuestionIf F(x)={3xx<1 find (a)f(2)(b)f(1)(c)f(c)F(x)=\left\{\begin{array}{lll}-3 x & x<-1 & \text { find }(a) f(-2)(b) f(-1)(c) f(c)\end{array}\right. (a) 3(2)=6-3(-2)=6 (b) 0 (c) 2(0)2+1=12(0)^{2}+1=1 What is the Domain

Studdy Solution

STEP 1

What is this asking? We're given a *piecewise function* F(x)F(x), which means it acts differently depending on the input xx, and we need to find what it spits out for a few specific xx values and state its *domain*, which are all the valid inputs. Watch out! Make sure to plug each xx into the *correct piece* of the function!
Also, don't mix up domain and range!

STEP 2

1. Evaluate F(2)F(-2)
2. Evaluate F(1)F(-1)
3. Evaluate F(0)F(0)
4. Determine the Domain

STEP 3

Since 2-2 is less than 1-1, we use the first piece of the function: F(x)=3xF(x) = -3x.

STEP 4

**Substitute** x=2x = -2 into the formula: F(2)=3(2)F(-2) = -3 \cdot (-2).

STEP 5

**Calculate** the result: F(2)=6F(-2) = 6.

STEP 6

For x=1x = -1, the second piece of the function applies: F(x)=0F(x) = 0.

STEP 7

Therefore, F(1)=0F(-1) = 0.
It's given to us directly!

STEP 8

Since 00 is greater than 1-1, we use the third piece: F(x)=2x2+1F(x) = 2x^2 + 1.

STEP 9

**Substitute** x=0x = 0 into the formula: F(0)=2(0)2+1F(0) = 2 \cdot (0)^2 + 1.

STEP 10

**Calculate** the result: F(0)=20+1=0+1=1F(0) = 2 \cdot 0 + 1 = 0 + 1 = 1.

STEP 11

The **domain** is all possible xx values we can plug into F(x)F(x).

STEP 12

The first piece covers x<1x < -1, the second piece covers x=1x = -1, and the third piece covers x>1x > -1.

STEP 13

Together, these pieces cover *all real numbers*!

STEP 14

(a) F(2)=6F(-2) = 6 (b) F(1)=0F(-1) = 0 (c) F(0)=1F(0) = 1 The domain of F(x)F(x) is all real numbers, which can be written as (,)(-\infty, \infty).

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