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
Chapter1: Expressions And Functions
Section: Chapter Questions
Problem 74SGR
Question
### 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.
\[ 
A'(4) = f(4) = 2(4) - 4 = 8 - 4 = 4. 
\]
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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