Question

1. the total number of patients that have been admitted to a hospital on a given day is given by a(t), where t is measured in hours since the beginning of the day (0 ≤ t ≤ 24). the rate at which patients are discharged or transferred out of the hospital, in patients per hour, is given by d(t). which of the following statements indicates that the number of patients currently in the hospital is decreasing?

a) a(t) - d(t) < 0
b) a(t) - d(t) < 0
c) a(t) - d(t) < 0
d) d(t) - a(t) < 0

1. the graph of y = f(x) is shown. the dashed line represents the line tangent to f at x = -1. what is the approximation of f(-1.1) using the line tangent to f at x = -1?

a) 1.0
b) 3.7
c) 3.9
d) 4.3
(image of the graph with points (-1, 4) and (0, 7) and the tangent line)

Explanation:

Question 4

Step1: Understand the rates

The function \( a(t) \) is the total number of admitted patients. The rate of admission is \( a'(t) \) (derivative of \( a(t) \) with respect to time \( t \)), and \( d(t) \) is the rate of discharge/transfer. The net rate of change of patients in the hospital is \( \text{Admission Rate} - \text{Discharge Rate} = a'(t) - d(t) \).

Step2: Determine decreasing condition

For the number of patients to be decreasing, the net rate of change must be negative. So we need \( a'(t) - d(t) < 0 \).

Step1: Find the equation of the tangent line

The tangent line at \( x = -1 \) passes through \( (-1, 4) \) and \( (0, 7) \). The slope \( m \) of the tangent line is \( \frac{7 - 4}{0 - (-1)} = \frac{3}{1} = 3 \). Using the point - slope form \( y - y_1 = m(x - x_1) \) with \( (x_1,y_1)=(-1,4) \), the equation of the tangent line is \( y - 4 = 3(x + 1) \), which simplifies to \( y = 3x+3 + 4=3x + 7 \).

Step2: Approximate \( f(-1.1) \)

We use the tangent line equation to approximate \( f(-1.1) \). Substitute \( x=-1.1 \) into \( y = 3x + 7 \): \( y=3(-1.1)+7=-3.3 + 7 = 3.7\)? Wait, no, wait. Wait, let's recalculate. Wait, \( 3\times(-1.1)=-3.3 \), \( -3.3 + 7 = 3.7\)? Wait, but let's check again. Wait, the tangent line passes through \( (-1,4) \) and \( (0,7) \). The slope is \( m=\frac{7 - 4}{0-(-1)} = 3 \). The equation is \( y=3(x + 1)+4=3x+3 + 4=3x + 7 \). Now, when \( x=-1.1 \), \( y = 3(-1.1)+7=-3.3 + 7 = 3.7\)? Wait, but let's check the options. Wait, maybe I made a mistake. Wait, the point \( (-1,4) \) and \( (0,7) \). The difference in \( x \) is \( 0-(-1)=1 \), difference in \( y \) is \( 7 - 4 = 3 \), so slope is 3. Then the equation is \( y - 4=3(x + 1) \), so \( y=3x+3 + 4=3x + 7 \). Now, \( x=-1.1 \), so \( y=3\times(-1.1)+7=-3.3 + 7 = 3.7\)? Wait, but option B is 3.7, option C is 3.9. Wait, maybe I messed up the points. Wait, the graph: the tangent line at \( x = - 1 \) (point \( (-1,4) \)) and \( (0,7) \). Wait, let's re - express the tangent line. Let's use the point \( (0,7) \). The equation is \( y=mx + b \), when \( x = 0 \), \( y = 7 \), so \( b = 7 \). The slope \( m = 3 \), so \( y=3x + 7 \). Now, for \( x=-1.1 \), \( y=3\times(-1.1)+7=-3.3 + 7 = 3.7\). Wait, but let's check the calculation again. \( 3\times(-1.1)=-3.3 \), \( 7-3.3 = 3.7\).

Answer:

A) \( a'(t) - d(t) < 0 \)

Question 5