Math  /  Algebra

Question-6 3. Given f(x)=x+3x5f(x)=\frac{x+3}{x-5} and g(x)=5xg(x)=\sqrt{5 x} find the following: XX (a) Does (fg)(5)(f \circ g)(5) exist? If so, find its' value. If not, explain why it does not exist. b) (fg)(x)(f \circ g)(x) 5(5)+35(5)5=1.78\frac{\sqrt{5(5)}+3}{\sqrt{5(5)}-5}=1.78 f(g(x))f(g(x)) 5x+35x5\frac{\sqrt{5 x}+3}{\sqrt{5 x}-5} c) (gf)(x)(g \circ f)(x) 5(x+3x5)\sqrt{5\left(\frac{x+3}{x-5}\right)} d) The domain of part bb E e) The domain of part c 12

Studdy Solution

STEP 1

What is this asking? We're playing with function composition!
We'll combine two functions, f(x)f(x) and g(x)g(x), in different ways and explore their domains. Watch out! Square roots can be tricky with domains, so let's be careful with what we plug in!

STEP 2

1. Evaluate the Composition at a Point
2. Compose f(g(x))
3. Compose g(f(x))
4. Find the Domain of f(g(x))
5. Find the Domain of g(f(x))

STEP 3

Let's **evaluate** g(5)g(5) first.
We have g(x)=5xg(x) = \sqrt{5x}, so g(5)=55=25=5g(5) = \sqrt{5 \cdot 5} = \sqrt{25} = 5.

STEP 4

Now, let's **plug** this result into f(x)f(x).
We have f(x)=x+3x5f(x) = \frac{x+3}{x-5}, so (fg)(5)=f(g(5))=f(5)=5+355=80(f \circ g)(5) = f(g(5)) = f(5) = \frac{5+3}{5-5} = \frac{8}{0}.

STEP 5

Uh oh!
We're dividing by **zero**, which means (fg)(5)(f \circ g)(5) is **undefined**!

STEP 6

To find (fg)(x)(f \circ g)(x), we **substitute** g(x)g(x) into f(x)f(x).
So, (fg)(x)=f(g(x))=f(5x)=5x+35x5(f \circ g)(x) = f(g(x)) = f(\sqrt{5x}) = \frac{\sqrt{5x}+3}{\sqrt{5x}-5}.

STEP 7

Now, let's find (gf)(x)(g \circ f)(x) by **substituting** f(x)f(x) into g(x)g(x).
We get (gf)(x)=g(f(x))=g(x+3x5)=5x+3x5=5x+15x5(g \circ f)(x) = g(f(x)) = g\left(\frac{x+3}{x-5}\right) = \sqrt{5 \cdot \frac{x+3}{x-5}} = \sqrt{\frac{5x+15}{x-5}}.

STEP 8

For the **domain** of (fg)(x)=5x+35x5(f \circ g)(x) = \frac{\sqrt{5x}+3}{\sqrt{5x}-5}, we need 5x05x \ge 0 for the square root, and 5x50\sqrt{5x} - 5 \ne 0 to avoid dividing by zero.

STEP 9

5x05x \ge 0 means x0x \ge 0. 5x50\sqrt{5x} - 5 \ne 0 means 5x5\sqrt{5x} \ne 5, so 5x255x \ne 25, and thus x5x \ne 5.

STEP 10

Combining these, the **domain** is x0x \ge 0 and x5x \ne 5, which we can write as [0,5)(5,)[0, 5) \cup (5, \infty).

STEP 11

For the **domain** of (gf)(x)=5x+15x5(g \circ f)(x) = \sqrt{\frac{5x+15}{x-5}}, the expression inside the square root must be non-negative.
So, we need 5x+15x50\frac{5x+15}{x-5} \ge 0.

STEP 12

We analyze the sign of the numerator and denominator.
The numerator, 5x+155x+15, is zero when x=3x=-3 and positive when x>3x > -3.
The denominator, x5x-5, is zero when x=5x=5 and positive when x>5x > 5.

STEP 13

The fraction is positive when both numerator and denominator have the same sign.
This happens when x3x \le -3 or x>5x > 5.
So, the **domain** is (,3](5,)(-\infty, -3] \cup (5, \infty).

STEP 14

a) (fg)(5)(f \circ g)(5) does **not exist** because it leads to division by zero. b) (fg)(x)=5x+35x5(f \circ g)(x) = \frac{\sqrt{5x}+3}{\sqrt{5x}-5}. c) (gf)(x)=5x+15x5(g \circ f)(x) = \sqrt{\frac{5x+15}{x-5}}. d) The domain of (fg)(x)(f \circ g)(x) is [0,5)(5,)[0, 5) \cup (5, \infty). e) The domain of (gf)(x)(g \circ f)(x) is (,3](5,)(-\infty, -3] \cup (5, \infty).

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