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Math  /  Calculus

QuestionThursday, November 28, 2024 Midterm Exam Calculus I d (0203101 \& 0213105 )
اكتب رمز الإجابة الصحيحة في الجدول بالحروف الكبيزة A, B, C, D \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline Question & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ \hline Answer & & & & & & & & & & & & & & & \\ \hline \end{tabular}
If f(x)=4x−1,g(x)=4xf(x)=\frac{4}{x-1}, g(x)=4 x, then the value of xx at which f∘g(x)=g∘f(x)f \circ g(x)=g \circ f(x) is: A) 14\frac{1}{4} B) 15\frac{1}{5} C) −13-\frac{1}{3} D) 13\frac{1}{3}
The graph of the function f(x)=(x2−7x+10)x2−25f(x)=\frac{\left(x^{2}-7 x+10\right)}{x^{2}-25} has at x=−5x=-5 A) jump B) Hole C) Vertical asymptote D) continuity point

Studdy Solution

STEP 1

1. We are given two functions: f(x)=4x−1 f(x) = \frac{4}{x-1} and g(x)=4x g(x) = 4x .
2. We need to find the value of x x where f∘g(x)=g∘f(x) f \circ g(x) = g \circ f(x) .
3. We are also given a function f(x)=x2−7x+10x2−25 f(x) = \frac{x^2 - 7x + 10}{x^2 - 25} and need to determine the behavior at x=−5 x = -5 .

STEP 2

1. Solve for f∘g(x) f \circ g(x) .
2. Solve for g∘f(x) g \circ f(x) .
3. Set f∘g(x)=g∘f(x) f \circ g(x) = g \circ f(x) and solve for x x .
4. Analyze the behavior of the function f(x)=x2−7x+10x2−25 f(x) = \frac{x^2 - 7x + 10}{x^2 - 25} at x=−5 x = -5 .

STEP 3

Calculate f∘g(x) f \circ g(x) :
f(g(x))=f(4x)=44x−1 f(g(x)) = f(4x) = \frac{4}{4x - 1}

STEP 4

Calculate g∘f(x) g \circ f(x) :
g(f(x))=g(4x−1)=4⋅4x−1=16x−1 g(f(x)) = g\left(\frac{4}{x-1}\right) = 4 \cdot \frac{4}{x-1} = \frac{16}{x-1}

STEP 5

Set f∘g(x)=g∘f(x) f \circ g(x) = g \circ f(x) and solve for x x :
44x−1=16x−1 \frac{4}{4x - 1} = \frac{16}{x-1}
Cross-multiply to solve:
4(x−1)=16(4x−1) 4(x-1) = 16(4x - 1) 4x−4=64x−16 4x - 4 = 64x - 16 12=60x 12 = 60x x=15 x = \frac{1}{5}

STEP 6

Analyze the behavior of f(x)=x2−7x+10x2−25 f(x) = \frac{x^2 - 7x + 10}{x^2 - 25} at x=−5 x = -5 :
The denominator x2−25=(x+5)(x−5) x^2 - 25 = (x+5)(x-5) is zero at x=−5 x = -5 , indicating a potential vertical asymptote. However, check the numerator:
The numerator x2−7x+10=(x−5)(x−2) x^2 - 7x + 10 = (x-5)(x-2) .
At x=−5 x = -5 , the numerator is not zero, confirming a vertical asymptote.
The value of x x for which f∘g(x)=g∘f(x) f \circ g(x) = g \circ f(x) is:
15 \boxed{\frac{1}{5}}
The graph of the function f(x)=x2−7x+10x2−25 f(x) = \frac{x^2 - 7x + 10}{x^2 - 25} has a:
Vertical asymptote \boxed{\text{Vertical asymptote}}

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