1) 2/7×1/2 2) 5/8×4/5 3) 1/6×2/34) 1/3×3/45) 1/2×2/36) 1/3×2/5
1 answer:
You might be interested in
Answer:
(B)
General Formulas and Concepts:
<u>Calculus</u>
Limits
Derivatives
- The definition of a derivative is the slope of the tangent line.
Derivative Notation
Instantaneous Rates
- Tangent Line:
Step-by-step explanation:
Since we are trying to find a <em>rate</em> at which W(t) changes, we must find the <em>derivative</em> at <em>t</em> = 3.
We are given 2 close answer choices that would have the same <em>numerical</em> answer but different <em>meanings</em>:
- (A)
- (B)
If we look at answer choice (A), we see that our units would simply just be volume. It would not have the units of a rate of change. Yes, it may be the closest numerically correct answer, but it does not tell us the <em>rate</em> at which the volume would be changing and it is not a derivative.
If we look at answer choice (B), we see that our units would be cm³/s, and that is most certainly a rate of change. Answer choice (B) is also a <em>derivative</em> at <em>t</em> = 3, and a derivative tells us what <em>rate</em> something is changing.
∴ Answer choice (B) will give us the best estimate for the value of the instantaneous rate of change of W(t) when <em>t</em> = 3.
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
Book: College Calculus 10e
Answer:
Where 0 < x < 3
The location of the local minimum, is (2, 0)
The location of the local maximum is at (0, 16)
Step-by-step explanation:
The given function is f(x) = (x + 2)⁴
The range of the minimum = 0 < x < 3
At a local minimum/maximum values, we have;
∴ (-x + 2)³ = 0
x = 2
When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2
We have, f(2) = (-2 + 2)⁴ = 0
The location of the local minimum, is (2, 0)
Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3
When x = 0, -x + 2 = 0 + 2 = 2
Similarly, we have;
-x + 2 = 1, when x = 1
-x + 2 = 0, when x = 2
-x + 2 = -1, when x = 3
Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16
The location of the local maximum is at (0, 16).