Exercise 82 Page 36

Practice makes perfect

a We are told that the cube root of x is -2 and asked to find the value of x. To do that, we can write the given equation with an exponent.

sqrt(x)=-2 ⇔ x=(-2)^3 Now, let's calculate the power.

x=(-2)^3

NegBaseToNegPow

(- a)^3 = - a^3

x=- 2^3

CalcPow

Calculate power

x=-8

b We are told that the square root of x is 12, and we are asked to find the value of x. To do that, we can write the given equation with an exponent.

sqrt(x)=12 ⇔ x=12^2 Now let's calculate the power.

x=12^2

PowToProdTwoFac

a^2=a* a

x=12*12

Multiply

x=144

c An absolute value measures a distance between an expression and zero.

|x+1|= 4This equation means that the distance is 4, either in the positive direction or the negative direction. |x+1|= 4 ⇒ lx+1= 4 x+1= -4 To find the solutions to the absolute value equation, we need to solve both of these cases for x.

| x+1|=4

lc x+1 ≥ 0:x+1 = 4 & (I) x+1 < 0:x+1 = - 4 & (II)

lcx+1=4 & (I) x+1=-4 & (II)

(I), (II): LHS-1=RHS-1

lx_1=3 x_2=-5

Both 3 and -5 are solutions to the absolute value equation.

Subchapter links

1.2.1 Describing a Graph p.19-21

1.2.2 Cube Root and Absolute Value Functions p.23-24

1.2.3 Function Machines p.27

1.2.4 Functions p.30-31

1.2.5 Domain and Range p.36

Exercise 82 Page 36

Hints

Check the answer

Practice exercises

Asses your skill