The fuel efficiency in miles per gallon for vehicles in the USA can be modelled by the function E(t) = -0.0007t^3 + 0.0278t^2 - 0.0843t + 12 where t is years since 1970, 0 less than or equal to t less than or equal to 35.       (a) Find E'(t)    (b) Fuel efficiency was improving (increasing) between the years a and b. Find the values of a and b.

### (a) Find \( E'(t) \)

The given function for fuel efficiency is:

\[
E(t) = -0.0007t^3 + 0.0278t^2 - 0.0843t + 12
\]

To find \( E'(t) \), we take the derivative of each term:

1. The derivative of \( -0.0007t^3 \) is \( -0.0007 \cdot 3t^2 = -0.0021t^2 \)
2. The derivative of \( 0.0278t^2 \) is \( 0.0278 \cdot 2t = 0.0556t \)
3. The derivative of \( -0.0843t \) is \( -0.0843 \)
4. The derivative of the constant \( 12 \) is \( 0 \)

Thus, the derivative \( E'(t) \) is:

\[
E'(t) = -0.0021t^2 + 0.0556t - 0.0843
\]

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### (b) Find the values of \( a \) and \( b \)

Fuel efficiency is increasing when \( E'(t) > 0 \). To find when this occurs, we need to solve \( E'(t) > 0 \).

The equation is:

\[
-0.0021t^2 + 0.0556t - 0.0843 > 0
\]

Let's solve this quadratic inequality.

First, solve the corresponding quadratic equation \( -0.0021t^2 + 0.0556t - 0.0843 = 0 \) to find the critical points.The critical points are approximately \( t = 1.61 \) and \( t = 24.86 \).

Now, to determine when fuel efficiency is increasing, we check the sign of \( E'(t) \) in the intervals determined by these critical points:

1. For \( t < 1.61 \)
2. For \( 1.61 < t < 24.86 \)
3. For \( t > 24.86 \)

We'll test the sign of \( E'(t) \) in these intervals.The results for \( E'(t) \) in the test intervals are:

- \( E'(t) < 0 \) for \( t < 1.61 \), indicating fuel efficiency is decreasing before \( t = 1.61 \).
- \( E'(t) > 0 \) for \( 1.61 < t < 24.86 \), indicating fuel efficiency is increasing between these values of \( t \).
- \( E'(t) < 0 \) for \( t > 24.86 \), indicating fuel efficiency is decreasing again after \( t = 24.86 \).

Thus, fuel efficiency was improving between the years:
- \( a = 1.61 \) (which corresponds to 1971.61, or approximately 1972)
- \( b = 24.86 \) (which corresponds to 1994.86, or approximately 1995).

### Final Answer:
- Fuel efficiency was improving between **1972** and **1995**.

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Would you like further details or explanations on any part of this? Here are 5 related questions you might find interesting:

1. How can we determine the maximum fuel efficiency within the interval?
2. What is the significance of \( E(t) = 0 \) in this context?
3. How can you solve more complex inequalities involving cubic or higher-degree functions?
4. What other real-world applications might use a similar mathematical model?
5. How does calculus help in optimizing real-life systems?

**Tip**: Quadratic inequalities can often be tested by checking the sign of the expression at various points in its intervals!

The fuel efficiency in miles per gallon for vehicles in the USA can be modelled by the function E(t) = -0.0007t^3 + 0.0278t^2 - 0.0843t + 12 where t is years since 1970, 0 <= t <= 35. (a) Find E'(t) (b) Fuel efficiency was improving (increasing) between the years a and b. Find the values of a and b.

Explore how to model and analyze fuel efficiency in vehicles using the function E(t) = -0.0007t^3 + 0.0278t^2 - 0.0843t + 12. Learn how to compute the derivative E'(t), solve quadratic inequalities, and find the years when fuel efficiency was improving. This problem involves key calculus concepts suitable for students in Grades 11-12 or early college.

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