Question

1. match each graph with its corresponding function.

1. let f(x) = 2x and g(x) = x^(2/3). find (fg)(x).

1. solve x + 2 = √(2x + 39).

the solution is x = □ .

1. let f(x) = 2x and g(x) = x^(2/3). find (fg)(x).

(fg)(x) = □

Explanation:

Problem 20: Solve \( x + 2 = \sqrt{2x + 39} \)

Step 1: Square both sides

To eliminate the square root, square both sides of the equation:
\( (x + 2)^2 = (\sqrt{2x + 39})^2 \)
Expanding the left side: \( x^2 + 4x + 4 = 2x + 39 \)

Step 2: Rearrange into quadratic equation

Subtract \( 2x + 39 \) from both sides:
\( x^2 + 4x + 4 - 2x - 39 = 0 \)
Simplify: \( x^2 + 2x - 35 = 0 \)

Step 3: Factor the quadratic

Find two numbers that multiply to \( -35 \) and add to \( 2 \). These are \( 7 \) and \( -5 \):
\( (x + 7)(x - 5) = 0 \)

Step 4: Solve for \( x \)

Set each factor to zero:
\( x + 7 = 0 \) or \( x - 5 = 0 \)
\( x = -7 \) or \( x = 5 \)

Step 5: Check solutions

• For \( x = -7 \): Left side \( -7 + 2 = -5 \), Right side \( \sqrt{2(-7) + 39} = \sqrt{25} = 5 \). \( -5

eq 5 \), so \( x = -7 \) is extraneous.

• For \( x = 5 \): Left side \( 5 + 2 = 7 \), Right side \( \sqrt{2(5) + 39} = \sqrt{49} = 7 \). \( 7 = 7 \), so \( x = 5 \) is valid.

Step 1: Recall function multiplication

The product of two functions \( (fg)(x) = f(x) \cdot g(x) \).

Step 2: Multiply the functions

Substitute \( f(x) = 2x \) and \( g(x) = x^{2/3} \):
\( (fg)(x) = 2x \cdot x^{2/3} \)
Use exponent rule \( a^m \cdot a^n = a^{m+n} \): \( 2x^{1 + 2/3} = 2x^{5/3} \) (or \( 2\sqrt[3]{x^5} \))

Answer:

\( x = 5 \)

Problem 21: Find \( (fg)(x) \) where \( f(x) = 2x \) and \( g(x) = x^{2/3} \)