Question

Consider the function

f(t)=1−2t+4t2.
Give the largest value of t such that the percentage rate of change equals 100. Give your answer to one decimal place.

Answer 1

To find the largest value of t such that the percentage rate of change equals 100, we need to find where the function's derivative is equal to 1. The largest value of t is approximately 0.4.

Step-by-step explanation:

To find the largest value of t such that the percentage rate of change equals 100, we need to find where the function's derivative is equal to 1.

The derivative of the function f(t) = 1 - 2t + 4t^2 is f'(t) = -2 + 8t.

Setting f'(t) equal to 1 and solving for t gives -2 + 8t = 1. Simplifying, we get 8t = 3, and dividing by 8 gives t = 0.375.

Therefore, the largest value of t such that the percentage rate of change equals 100 is approximately 0.4.

Answer 2

The percentage rate of change of a function f(t) is given by:

(Δf / f) x 100

where Δf is the change in f and f is the original value of the function.

To find the largest value of t such that the percentage rate of change equals 100, we need to find the value of t for which:

(Δf / f) x 100 = 100

Simplifying, we get:

Δf / f = 1

This means that the change in f is equal to the original value of f.

So, we need to solve the equation:

f(t + Δt) - f(t) = f(t)

where Δt is the change in t.

Substituting the given function, we get:

[1 - 2(t + Δt) + 4(t + Δt)^2] - [1 - 2t + 4t^2] = 1 - 2t + 4t^2

Simplifying, we get:

-8tΔt + 8Δt^2 = 1

Since we are interested in the largest value of t, we can assume that Δt is a small positive number, such that Δt << t.

Ignoring the term Δt^2, we get:

-8tΔt = 1

Solving for Δt, we get:

Δt = -1 / (8t)

Substituting this value of Δt back into the equation -8tΔt + 8Δt^2 = 1, we get:

-8t(-1 / (8t)) + 8(-1 / (8t))^2 = 1

Simplifying, we get:

1 / t^2 = 1

Solving for t, we get:

t = 1

Therefore, the largest value of t such that the percentage rate of change equals 100 is t = 1.