Mathway | Alg Do Homework Not syncing https://www.mathway.com MATH-1325-006 Homework: HW 11.3 Score: 0 of 1 pt 7 of 9 (7 complete) Save HW Score: 55 Question Help 11.3.17 Find the instantaneous rate of change for the function at the given value: g(t) = 1 – √(t). The instantaneous rate of change at t = -2 is
The Correct Answer and Explanation is:
Based on the image provided, here is the correct answer to the math problem and a detailed explanation.
Correct Answer: 4
Explanation
The problem asks for the instantaneous rate of change of the function g(t) = 1 – t² at the specific value t = -2. In calculus, the instantaneous rate of change of a function at a point is found by calculating the derivative of the function and then evaluating it at that specific point. The derivative represents the slope of the tangent line to the function’s graph at that point.
The process involves two main steps:
Step 1: Find the derivative of the function g(t).
The given function is g(t) = 1 – t². To find its derivative, which we’ll call g'(t), we differentiate the function term by term with respect to t.
- Derivative of the constant term: The first term is 1, which is a constant. The derivative of any constant is always zero.
- Derivative of the variable term: The second term is -t². To differentiate this, we use the power rule, which states that the derivative of tⁿ is ntⁿ⁻¹. In this case, n=2. So, the derivative of t² is 2t²⁻¹ = 2t. Since our term is -t², its derivative is -2t.
Combining the derivatives of both terms, we get:
g'(t) = 0 – 2t
g'(t) = -2t
This derivative function, g'(t) = -2t, gives us the formula for the instantaneous rate of change of g(t) at any value of t.
Step 2: Evaluate the derivative at the given value, t = -2.
Now that we have the derivative, we can find the instantaneous rate of change at t = -2 by substituting -2 for t in the derivative function g'(t).
g'(-2) = -2 * (-2)
Multiplying the two negative numbers gives a positive result:
g'(-2) = 4
Therefore, the instantaneous rate of change of the function g(t) = 1 – t² at t = -2 is 4. This positive value indicates that at the precise point where t = -2, the function is increasing at a rate of 4 units of g(t) per unit of t.
