Math Problem Statement

Question content area top Part 1 For f left parenthesis x right parenthesis equals StartFraction 9 Over x plus 6 EndFraction and g left parenthesis x right parenthesis equals StartFraction 1 Over x EndFraction ​, find a.  left parenthesis f circle g right parenthesis left parenthesis x right parenthesis​;     b.  the domain of f circle g Question content area bottom Part 1 a.  left parenthesis f circle g right parenthesis left parenthesis x right parenthesisequals    StartFraction 9 x Over 1 plus 6 x EndFraction ​(Simplify your​ answer.) Part 2 b.  What is the domain of f circle g​? The domain is    left parenthesis negative infinity comma negative one sixth right parenthesis union left parenthesis negative one sixth comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis .
​(Simplify your answer. Type your answer in interval notation. Use integers or fractions for any numbers in the​ expression.)

Solution

We are given two functions f(x)=9x+6f(x) = \frac{9}{x+6} and g(x)=1xg(x) = \frac{1}{x}. We need to find:

  1. (fg)(x)(f \circ g)(x) (which means f(g(x))f(g(x))).
  2. The domain of fgf \circ g.

Part 1: Find (fg)(x)(f \circ g)(x)

To find (fg)(x)(f \circ g)(x), we need to substitute g(x)g(x) into f(x)f(x):

(fg)(x)=f(g(x))=f(1x)(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right)

Now, substitute 1x\frac{1}{x} for xx in f(x)f(x):

f(1x)=91x+6f\left(\frac{1}{x}\right) = \frac{9}{\frac{1}{x} + 6}

Simplify the expression:

f(1x)=91x+6=91+6xx=9x1+6xf\left(\frac{1}{x}\right) = \frac{9}{\frac{1}{x} + 6} = \frac{9}{\frac{1 + 6x}{x}} = \frac{9x}{1 + 6x}

So,

(fg)(x)=9x1+6x(f \circ g)(x) = \frac{9x}{1 + 6x}

Part 2: Domain of fgf \circ g

To find the domain of fgf \circ g, we need to consider the restrictions from both f(x)f(x) and g(x)g(x).

  • For g(x)=1xg(x) = \frac{1}{x}, the function is undefined when x=0x = 0 (since division by zero is not allowed). Therefore, x0x \neq 0.
  • For f(x)=9x+6f(x) = \frac{9}{x+6}, the function is undefined when x+6=0x+6 = 0, or x=6x = -6.

Now, since f(g(x))=9x1+6xf(g(x)) = \frac{9x}{1 + 6x}, we have to ensure that the denominator 1+6x01 + 6x \neq 0, which gives x16x \neq -\frac{1}{6}.

Thus, the domain excludes x=0x = 0 and x=16x = -\frac{1}{6}.

Therefore, the domain of fgf \circ g is all real numbers except 0 and 16-\frac{1}{6}, written in interval notation as:

Domain of fg:(,16)(16,0)(0,)\text{Domain of } f \circ g: (-\infty, -\frac{1}{6}) \cup (-\frac{1}{6}, 0) \cup (0, \infty)


Would you like a more detailed explanation of any part? Here are 5 related questions for practice:

  1. What is the domain of f(x)=9x+6f(x) = \frac{9}{x+6}?
  2. Find (gf)(x)(g \circ f)(x) for f(x)=9x+6f(x) = \frac{9}{x+6} and g(x)=1xg(x) = \frac{1}{x}.
  3. How do you determine the domain of a composition of two functions?
  4. Can a function have more than one domain restriction?
  5. Why is x=0x = 0 excluded from the domain of g(x)=1xg(x) = \frac{1}{x}?

Tip: Always simplify the composition of functions before determining the domain to avoid unnecessary restrictions.

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Math Problem Analysis

Mathematical Concepts

Function Composition
Rational Functions
Domain of Functions

Formulas

(f ∘ g)(x) = f(g(x))
f(x) = 9/(x + 6)
g(x) = 1/x

Theorems

Domain Restrictions in Rational Functions

Suitable Grade Level

Grades 10-12