Question

1. Describe the graph of the function without drawing the graph. Use the terms increasing, decreasing, concave up, concave down, vertical intercept, and horizontal asymptote, as appropriate.

f(t) = 0.75(3)^t − 2

.the vertical intercept is (t, f(t)) =

.The function is increasing or decreasing and concave up or concave down?

2. Model the data with an exponential function, if appropriate. If an exponential function model is not appropriate for the situation, explain why. Do not use regression for this exercise.

Between 1960 and 2004, insurance company expenditures for health care increased at an ever-increasing rate. In 1960, $6 billion was spent on health care. In 2004, $659 billion was spent on health care.†

Is the exponential function model appropriate? If not, explain why.

. Yes, the exponential function model is appropriate.

. No, the exponential function model is not appropriate because there is a constant rate of change, meaning it is linear.  

. No, the exponential function model is not appropriate because the growth rate is less than 1.

. No, the exponential function model is not appropriate because the change factor is less than 1.

Give the exponential model, if appropriate. (If not appropriate, enter DNE. Let H(t) be health care expenditures by insurance companies in billions of dollars, t years after 1960. Round the change factor to two decimal places.)

H(t) = ________ billion dollars.

3. State the two integer values between which the expression falls.

log₂(40)

Smaller value:-

Larger value:-

Use the change of base formula to evaluate the expression exactly. (Round your answer to three decimal places.):- __________

Answer 1

(1)

Consider the function.

f(t) = 0.75(3)^t − 2 ......(1)

For the vertical intercept, putting t =0 in equation (1).

f(t) = 0.75(3)⁰ -2 = 0.75(1) - 2 = 0.75 -2 = -1.25

So, vertical intercept is (t, f(t)) = (0, -1.25)

To check whether the function is increasing or decreasing and concave up or concave down.

Let differentiate the function wrt t to get

Differentiate equation (1) to get

.......(2)

Here, f'(t) >0 for any values of t.

Since f'(x) is always positive. So, the function always increasing.

So, f(x) is increasing at t = (-∞, ∞)

Now,

Again differentiate equation (2) wrt t to get

This is also greater than 0.

So, f"(x) is also positive.

Since both f'(x) and f"(x) are positive. So, the graph of function is concave up.

Again consider the equation of the function

f(t) = 0.7 5(3)^t -2

We can observe that as t is tending to -∞, f(t) → -2.

So, y =-2 is horizontal asymptote.

(2)

Let an exponential function representing health expenditure as a function of time in t years after year 1960 be

H(t) = a(b)^t ....(3)

Here, a = initial (in 1960) health expenditure.

And b = growth rate

Now, since in 1960, $6 billion was spent on health care.

On putting a =6 in equation (3) we get

H(t) = 6(b)^t .....(4)

Since in 2004, $659 billion was spent on health cares .

So, for t = 2004 -1960 = 44 years H(t) = 659 billion.

On putting these values in equation (4), we get

659 = 6(b)⁴⁴

On reversing the sides we get

6(b)⁴⁴ = 659

Or (b)⁴⁴ = 659/6

Or b⁴⁴ = 109.83

Or b = (109.33)^1÷44

Or b = 1.1127 ≈ 1.11

On putting this value of b in equation (4), we get

H(t) = 6(1.11)^t

Yes, the exponential function model appropriate because growth rate is greater than 1.

Hence, H(t) health care expenditures by insurance companies in billions of dollars, t years after 1960 is

H(t) = 6(1.11)^t billion dollars.

(3)

log₂(40) is lies between

log₂(32) and log₂(64)

Or log₂(2⁵) and log₂(2⁶)

Or 5 and 6

(Because log_a(a)ⁿ = n)

Hence, the two integer values between which the expression log₂(40) falls are

Smaller value:- 5

Larger value:- 6

Now, using the change of base formula to evaluate the expression exactly.

Since log_a(b) = log₁₀(b) / log₁₀(a)

So,

log₂(40)

= log₁₀(40) / log₁₀(2)

=1.60206 - 0.301026

=1.301

Hence, log₂(40) = 1.301