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lesya [120]
3 years ago
10

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 per day, c, consumed by the subject. In terms of diet and weight, interpret the statement f(1850) = 158.
a. For this person, decreasing calorie intake by 1850 Calories per day will make his/her weight decrease to 158 pounds.
b. For this person, increasing calorie intake by 1850 Calories per day will make his/her weight decrease to 158 pounds.
c. For this person, consuming 1850 Calories per day results in a weight of 158 pounds.
d. For this person, decreasing calorie intake by 1850 Calories per day will make his/her weight increase to 158 pounds.
e. For this person, increasing calorie intake by 1850 Calories per day will make his/her weight increase to 158 pounds
Mathematics
1 answer:
Tju [1.3M]3 years ago
3 0

Answer: Option C

Step-by-step explanation:

From the given condition, W(weight) is a function of C(calories), that is, the value of W is highly dependent on the value of C. In this case, C, is the independent variable and W is the dependent variable. Option C is more correct than other options because the question did not state if increasing C or decreasing C means we havw an increasing W or decreasing W, it only stated that W is. function of C.

Hence when an average of 1850 of calories is consumed, the resulting weight is 158pounds.

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Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

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