Math Problem Statement

A laboratory study investigating the relationship between diet and weight in adult humans found that the weight of a subject, W, in pounds, was a function, W = f(c), of the average number of Calories, c, consumed by the subject in a day.

Solution

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The problem involves the function W=f(c)W = f(c), where WW represents weight in pounds, and cc represents the number of calories consumed by the subject in a day.

Here's a breakdown of each part:

(a) In the statement f(1900)=160f(1900) = 160

  • The function f(1900)f(1900) means the weight WW of a subject when they consume 1900 calories is 160 pounds.
  • The units of 1900 are calories (cal).
  • The units of 160 are pounds (lb).

(b) In the statement f(2000)=165f(2000) = 165

  • The units of 2000 are also calories (cal), as it's a daily calorie intake.
  • The units of 165 are pounds (lb), as this refers to the weight of the subject.

(c) In the statement f1(169)=2200f^{-1}(169) = 2200

  • This inverse function means that if the subject's weight is 169 pounds, the subject is consuming 2200 calories per day.
  • The units of 169 are pounds (lb), as this is the weight.
  • The units of 2200 are calories (cal).

(d) What are the units of f(c)=dWdcf'(c) = \frac{dW}{dc}?

  • f(c)f'(c) represents the rate of change of weight with respect to calories consumed.
  • The units of f(c)f'(c) are pounds per calorie (lb/cal).

(e) If Sam increases her calorie intake from 2600 to 2620 calories, her weight will increase by approximately 0.2 pounds.

  • Here you need to find f(2600)f'(2600).
  • The weight change per calorie is f(2600)f'(2600), so 0.2 pounds=f(2600)×(26202600)0.2 \text{ pounds} = f'(2600) \times (2620 - 2600).
  • Solving for f(2600)f'(2600), you get: f(2600)=0.2 lb20 cal=0.01 lb/cal.f'(2600) = \frac{0.2 \text{ lb}}{20 \text{ cal}} = 0.01 \text{ lb/cal}.

This fills the blank for f(2600)f'(2600), which is 0.01 lb/cal.


Would you like further details on any part or have any other questions?

Five Related Questions:

  1. How is the derivative f(c)f'(c) interpreted in terms of weight change with calorie intake?
  2. What would the weight change be if Sam increased her intake from 2500 to 2600 calories?
  3. Can we determine Sam’s weight if she consumes 1900 calories a day based on this model?
  4. What does f1(W)f^{-1}(W) tell us about a person's calorie intake if we know their weight?
  5. How would the interpretation change if we used a different measurement unit for weight (e.g., kilograms)?

Tip:

When interpreting inverse functions like f1(x)f^{-1}(x), remember they reverse the input-output relationship, letting you find inputs from known outputs.

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Math Problem Analysis

Mathematical Concepts

Functions
Derivatives
Inverse Functions

Formulas

f(1900) = 160
f(2000) = 165
f^{-1}(169) = 2200
f'(c) = dW/dc
f'(2600) = 0.01 lb/cal

Theorems

Derivative of a function
Inverse function theorem

Suitable Grade Level

Grades 11-12