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question 10 of 15 two functions g and f are defined in the figure below…

Question

question 10 of 15
two functions g and f are defined in the figure below.

find the domain and range of the composition $f \circ g$. write your answers in set notation.
(a) domain of $f \circ g: $
(b) range of $f \circ g: $
question 11 of 15
for the real-valued functions $g(x)=\frac{x-4}{x-1}$ and $h(x)=4x-11$, find the composition $g \circ h$. also, specify
its domain using interval notation.
$(g \circ h)(x)=$
domain of $g \circ h:$
for the real-valued functions $g(x)=\frac{x-4}{x-1}$ and $h(x)=4x-11$, find the composition $g \circ h$. also, specify
its domain using interval notation.
$g \circ h\\ x =$
domain of $g \circ h:$

Explanation:

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Question 10

Step1: Identify valid domain for $f\circ g$

The composition $f\circ g(x) = f(g(x))$ requires $g(x)$ to be in the domain of $f$. Check each input in domain of $g$:

  • $g(1)=9$: 9 is not in domain of $f$ → invalid
  • $g(2)=7$: 7 is in domain of $f$ → valid
  • $g(4)=0$: 0 is in domain of $f$ → valid
  • $g(5)=4$: 4 is not in domain of $f$ → invalid
  • $g(6)=5$: 5 is in domain of $f$ → valid
  • $g(8)=8$: 8 is in domain of $f$ → valid

Valid inputs: $\{2,4,6,8\}$

Step2: Calculate $f\circ g$ for valid inputs

Compute each valid composition:

  • $f(g(2))=f(7)=7$
  • $f(g(4))=f(0)=0$
  • $f(g(6))=f(5)=6$
  • $f(g(8))=f(8)=8$

Resulting outputs: $\{0,6,7,8\}$

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Question 11

Step1: Substitute $h(x)$ into $g(x)$

$g\circ h(x) = g(h(x))$, substitute $h(x)=4x-11$ into $g(t)=\frac{t-4}{t-1}$:
$$g(h(x)) = \frac{(4x-11)-4}{(4x-11)-1}$$

Step2: Simplify the expression

Simplify numerator and denominator:
Numerator: $(4x-11)-4 = 4x-15$
Denominator: $(4x-11)-1 = 4x-12 = 4(x-3)$
$$g\circ h(x) = \frac{4x-15}{4x-12} = \frac{4x-15}{4(x-3)}$$

Step3: Find domain restrictions
  1. Denominator of $g\circ h$ cannot be 0:

$4x-12
eq 0 \implies x
eq 3$

  1. $h(x)$ must be in domain of $g$ (denominator of $g$ cannot be 0):

$h(x)-1
eq 0 \implies 4x-11-1
eq 0 \implies 4x-12
eq 0 \implies x
eq 3$
Combine restrictions: $x \in (-\infty, 3) \cup (3, \infty)$

Answer:

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Question 10

(a) Domain of $f\circ g$: $\{2, 4, 6, 8\}$
(b) Range of $f\circ g$: $\{0, 6, 7, 8\}$

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Question 11

$(g\circ h)(x) = \frac{4x-15}{4x-12}$
Domain of $g\circ h$: $(-\infty, 3) \cup (3, \infty)$