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Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

3. Function Notation

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Exercise 18 Page 125

Substitute -5 for j(x).

15

Practice makes perfect

To find the value of x that will make the function equal to -5, we will substitute -5 for j(x) in the given function rule. Then, we can solve for x.

j(x)=-4/5x+7

j(x)= -5

-5=-4/5x+7

▼

Solve for x

LHS-7=RHS-7

-5- 7=-4/5x+7- 7

Subtract terms

-12=-4/5x

MoveRightFacToNum

a/c* b = a* b/c

-12=-4x/5

LHS * (-1)=RHS* (-1)

12=4x/5

LHS * 5=RHS* 5

12 * 5=4x/5 * 5

Multiply

60=4x/5 * 5

FracMultDenomToNumber

a/5* 5 = a

60=4x

.LHS /4.=.RHS /4.

60/4=4x/4

Calculate quotient

15=4x/4

MovePartNumRight

a* b/c=a/c* b

15=4/4 * x

a/a=1

15=1 * x

Identity Property of Multiplication

15=x

Rearrange equation

x=15

When x=15, the function equals -5.

Checking Our Answer

Verifying Our Solution

We can verify our solution by substituting x=15 into the equation and evaluating.

j(x)=-4/5x+7

x= 15

j( 15)=-4/5( 15)+7

MoveRightFacToNum

a/c* b = a* b/c

j(15)=-4 * 15/5+7

Multiply

j(15)=-60/5+7

Calculate quotient

j(15)=-12+7

Add terms

j(15)=-5

When x=15, j(x)=-5, so our solution is correct.

Subchapter links

Communicate Your Answer p.121

Monitoring Progress p.122-124

Exercises p.125-126