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Find the Relative Change in the values of the function f(x) = x3 + 2 between x = 1 and x = 2.5 Enter your answer accurate to three digits to the right of the decimal point.

User Oldergod
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2 Answers

3 votes

Final answer:

The relative change in the values of the function f(x) = x^3 + 2 between x = 1 and x = 2.5 is 4.875.

Step-by-step explanation:

To find the relative change in the values of the function f(x) = x3 + 2 between x = 1 and x = 2.5, we first calculate the values of the function at these points:

  • For x = 1: f(1) = 13 + 2 = 1 + 2 = 3
  • For x = 2.5: f(2.5) = 2.53 + 2 = 15.625 + 2 = 17.625

Next, we calculate the absolute change and then the relative change.

  1. Absolute change = f(2.5) - f(1) = 17.625 - 3 = 14.625
  2. Relative change = Absolute change / f(1) = 14.625 / 3 ≈ 4.875

The relative change, accurate to three digits to the right of the decimal point, is 4.875.

User David Bendory
by
7.8k points
4 votes

Answer:

9.750

Step-by-step explanation:

formula for average rate of change is ;

f(b)- f(a)/b- a

here a = 1 and b = 2.5

f(b) = (2.5)^3 + 2 = 17.625

f(a) = 1^3+ 2 = 3

using the formula above, we have it that

(17.625-3)/2.5-1 = 9.750

User Mochi
by
8.2k points

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