Math  /  Algebra

QuestionJohn is walking. D(t)D(t), given below, is his distance in kilometers from Tomsville after tt hours of walking. D(t)=15.55tD(t)=15.5-5 t
Complete the following statements.
Let D1D^{-1} be the inverse function of DD. Take xx to be an output of the function DD. That is, x=D(t)x=D(t) and t=D1(x)t=D^{-1}(x). (a) Which statement best describes D1(x)D^{-1}(x) ? The reciprocal of his distance from Tomsville (in kilometers) after walking xx hours. The amount of time he has walked (in hours) when he is xx kilometers from Tomsville. His distance from Tomsville (in kilometers) after he has walked xx hours. The ratio of the amount of time he has walked (in hours) to his distance from Tomsville (in kilometers). xt. x_{\text {t. }} (b) D1(x)=D^{-1}(x)= \square (c) D1(8:5)=D^{-1}(8: 5)= \square

Studdy Solution

STEP 1

What is this asking? Given a formula for John's distance from Tomsville based on how long he's been walking, we need to figure out how long it takes him to reach a certain distance, and then use that to find how long it takes him to reach 8.5 kilometers. Watch out! Don't mix up distance and time! D(t)D(t) gives distance based on time, but D1(x)D^{-1}(x) gives time based on distance.
Also, be super careful with your calculations!

STEP 2

1. Describe the inverse function
2. Find the inverse function
3. Calculate the time

STEP 3

D(t)D(t) tells us John's **distance** after walking for tt **hours**.
So, D1(x)D^{-1}(x) must tell us the **time** it takes him to be xx **kilometers** away from Tomsville.

STEP 4

That means the correct description for D1(x)D^{-1}(x) is "The amount of time he has walked (in hours) when he is xx kilometers from Tomsville."

STEP 5

We're given D(t)=15.55tD(t) = 15.5 - 5t, and we know that x=D(t)x = D(t).
So, let's write x=15.55tx = 15.5 - 5t.

STEP 6

Now, we want to **solve for** tt in terms of xx.
This will give us D1(x)D^{-1}(x).
First, let's subtract 15.5 from both sides of the equation: x15.5=15.55t15.5x - 15.5 = 15.5 - 5t - 15.5, which simplifies to x15.5=5tx - 15.5 = -5t.

STEP 7

Next, let's **divide both sides** by -5 to isolate tt: x15.55=5t5\frac{x - 15.5}{-5} = \frac{-5t}{-5}.
This gives us t=x15.55t = \frac{x - 15.5}{-5}.
We can rewrite this as t=15.5x5t = \frac{15.5 - x}{5} to avoid the negative sign in the denominator.

STEP 8

So, D1(x)=15.5x5D^{-1}(x) = \frac{15.5 - x}{5}.
This formula tells us the **time** (in hours) it takes John to be xx kilometers from Tomsville.

STEP 9

We want to find D1(8.5)D^{-1}(8.5), which represents the time it takes John to be 8.5 kilometers from Tomsville.
We just plug in x=8.5x = 8.5 into our inverse function: D1(8.5)=15.58.55D^{-1}(8.5) = \frac{15.5 - 8.5}{5}.

STEP 10

Now, we **subtract** in the numerator: D1(8.5)=75D^{-1}(8.5) = \frac{7}{5}.

STEP 11

Finally, we **divide**: D1(8.5)=1.4D^{-1}(8.5) = 1.4.

STEP 12

(a) The amount of time he has walked (in hours) when he is xx kilometers from Tomsville. (b) D1(x)=15.5x5D^{-1}(x) = \frac{15.5 - x}{5} (c) D1(8.5)=1.4D^{-1}(8.5) = 1.4

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