5. Let f (t) = 2t – 4. %3D А. Sketch or provide the graph of the function f on the interval [0,5]. Given A(x) = S ƒ(t)dt, evaluate the following. Explain: i. A(4) ii. А(3) iii. A' (4) B.
5. Let f (t) = 2t – 4. %3D А. Sketch or provide the graph of the function f on the interval [0,5]. Given A(x) = S ƒ(t)dt, evaluate the following. Explain: i. A(4) ii. А(3) iii. A' (4) B.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter3: Linear And Nonlinear Functions
Section: Chapter Questions
Problem 26MCQ
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![### Problem 5
**Let \( f(t) = 2t - 4 \).**
**A.** Sketch or provide the graph of the function \( f \) on the interval \([0, 5]\).
*Explanation of Graph*: The function \( f(t) = 2t - 4 \) is a linear function with a slope of 2 and a y-intercept of -4. On the interval \([0, 5]\), at \( t = 0 \), \( f(t) = -4 \). At \( t = 5 \), \( f(t) = 2(5) - 4 = 6 \). So, the graph is a straight line starting from the point (0, -4) to (5, 6).
**B.** Given \( A(x) = \int_{3}^{x} f(t) \, dt \), evaluate the following. Explain:
**i.** \( A(4) \)
**Explanation**: Calculate the definite integral of \( f(t) = 2t - 4 \) from 3 to 4:
\[
A(4) = \int_{3}^{4} (2t - 4) \, dt.
\]
Solve the integral to find the area under the curve from \( t = 3 \) to \( t = 4 \).
**ii.** \( A(3) \)
**Explanation**: Since the integral is evaluated from 3 to 3:
\[
A(3) = \int_{3}^{3} (2t - 4) \, dt = 0.
\]
The area under the curve from a point to itself is zero.
**iii.** \( A'(4) \)
**Explanation**: According to the Fundamental Theorem of Calculus, the derivative of the integral function with respect to x, \( A'(x) \), at a point is equal to the value of the original function \( f(t) \) at that point.
\[
A'(4) = f(4) = 2(4) - 4 = 8 - 4 = 4.
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1b542d2-f99f-41f7-9f8d-fbf0665f8261%2Fa79f1811-5b3a-41f7-b396-ca69d0cac846%2Fnpuvt5_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 5
**Let \( f(t) = 2t - 4 \).**
**A.** Sketch or provide the graph of the function \( f \) on the interval \([0, 5]\).
*Explanation of Graph*: The function \( f(t) = 2t - 4 \) is a linear function with a slope of 2 and a y-intercept of -4. On the interval \([0, 5]\), at \( t = 0 \), \( f(t) = -4 \). At \( t = 5 \), \( f(t) = 2(5) - 4 = 6 \). So, the graph is a straight line starting from the point (0, -4) to (5, 6).
**B.** Given \( A(x) = \int_{3}^{x} f(t) \, dt \), evaluate the following. Explain:
**i.** \( A(4) \)
**Explanation**: Calculate the definite integral of \( f(t) = 2t - 4 \) from 3 to 4:
\[
A(4) = \int_{3}^{4} (2t - 4) \, dt.
\]
Solve the integral to find the area under the curve from \( t = 3 \) to \( t = 4 \).
**ii.** \( A(3) \)
**Explanation**: Since the integral is evaluated from 3 to 3:
\[
A(3) = \int_{3}^{3} (2t - 4) \, dt = 0.
\]
The area under the curve from a point to itself is zero.
**iii.** \( A'(4) \)
**Explanation**: According to the Fundamental Theorem of Calculus, the derivative of the integral function with respect to x, \( A'(x) \), at a point is equal to the value of the original function \( f(t) \) at that point.
\[
A'(4) = f(4) = 2(4) - 4 = 8 - 4 = 4.
\]
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