Math  /  Algebra

QuestionSuppose that the functions ff and gg are defined as follows. f(x)=4x5g(x)=5x+1\begin{array}{l} f(x)=4 x-5 \\ g(x)=\sqrt{5 x+1} \end{array}
Find fgf \cdot g and fgf-g. Then, give their domains using interval notation. (fg)(x)=(f \cdot g)(x)= \square
Domain of fgf \cdot g : \square (fg)(x)=(f-g)(x)= \square
Domain of fgf-g : \square

Studdy Solution

STEP 1

What is this asking? We're given two functions, f(x)f(x) and g(x)g(x), and we need to find fgf \cdot g (that's ff **times** gg), and fgf - g (that's ff **minus** gg).
Then, we need to figure out what input values, or xx values, are allowed for each of those new combined functions. Watch out! Remember that when we combine functions, the domain can change!
We need to consider the domains of the original functions *and* any new restrictions created by combining them.

STEP 2

1. Find fgf \cdot g.
2. Find the domain of fgf \cdot g.
3. Find fgf - g.
4. Find the domain of fgf - g.

STEP 3

To find fgf \cdot g, we **multiply** the expressions for f(x)f(x) and g(x)g(x) together.
It's as simple as that! (fg)(x)=f(x)g(x)=(4x5)(5x+1)(f \cdot g)(x) = f(x) \cdot g(x) = (4x - 5) \cdot (\sqrt{5x + 1})

STEP 4

In this case, we can't really simplify further, so we leave it as: (fg)(x)=(4x5)5x+1(f \cdot g)(x) = (4x - 5)\sqrt{5x + 1}

STEP 5

The function f(x)=4x5f(x) = 4x - 5 is a linear function, and linear functions are defined for all real numbers.
So, the domain of f(x)f(x) is all real numbers.

STEP 6

The function g(x)=5x+1g(x) = \sqrt{5x + 1} has a square root.
We know that we can't take the square root of a negative number, so we need 5x+15x + 1 to be greater than or equal to zero: 5x+105x + 1 \ge 0 Subtract 1 from both sides: 5x15x \ge -1 Divide both sides by 5: x15x \ge -\frac{1}{5} So, the domain of g(x)g(x) is [15,)\left[ -\frac{1}{5}, \infty \right).

STEP 7

Since we need *both* f(x)f(x) *and* g(x)g(x) to be defined, we take the **intersection** (overlap) of their domains.
The domain of f(x)f(x) is all real numbers, and the domain of g(x)g(x) is x15x \ge -\frac{1}{5}.
The overlap is just x15x \ge -\frac{1}{5}, so the domain of fgf \cdot g is [15,)\left[ -\frac{1}{5}, \infty \right).

STEP 8

To find fgf - g, we **subtract** g(x)g(x) from f(x)f(x): (fg)(x)=f(x)g(x)=(4x5)(5x+1)(f - g)(x) = f(x) - g(x) = (4x - 5) - (\sqrt{5x + 1})

STEP 9

This can't be simplified further, so: (fg)(x)=4x55x+1(f - g)(x) = 4x - 5 - \sqrt{5x + 1}

STEP 10

Just like with fgf \cdot g, we need to consider the domains of f(x)f(x) and g(x)g(x).
We already know that the domain of f(x)f(x) is all real numbers and the domain of g(x)g(x) is x15x \ge -\frac{1}{5}.

STEP 11

Again, we take the **intersection** of the domains, which is x15x \ge -\frac{1}{5}.
So, the domain of fgf - g is [15,)\left[ -\frac{1}{5}, \infty \right).

STEP 12

(fg)(x)=(4x5)5x+1(f \cdot g)(x) = (4x - 5)\sqrt{5x + 1}
Domain of fgf \cdot g: [15,)\left[ -\frac{1}{5}, \infty \right)
(fg)(x)=4x55x+1(f - g)(x) = 4x - 5 - \sqrt{5x + 1}Domain of fgf - g: [15,)\left[ -\frac{1}{5}, \infty \right)

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