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The question asks for the approximate value of Δy as x varies from 1 to 2, given the function \( f(x) = 1.4 - 0.5x \) and Δx = 0.5.

To solve for Δy, we use the formula:

\[
\Delta y \approx f'(x) \cdot \Delta x
\]

First, we need to find the derivative \( f'(x) \) of the function \( f(x) = 1.4 - 0.5x \):

\[
f'(x) = -0.5
\]

Now, using \( f'(x) = -0.5 \) and Δx = 0.5:

\[
\Delta y \approx -0.5 \times 0.5 = -0.25
\]

So, the approximate value of Δy is **-0.25**.

Would you like further details or have any questions?

Here are 5 related questions for practice:
1. How do you find the derivative of a linear function?
2. If Δx was doubled, how would it affect Δy?
3. How does the slope of a function affect the sign of Δy?
4. How do you approximate changes in non-linear functions?
5. What is the importance of using the derivative for approximation?

**Tip:** The derivative tells you the rate of change of a function, and multiplying it by a small change in \( x \) (Δx) gives you an approximation for the change in \( y \) (Δy).

If x = 1 and Δx = 0.5, what is the approximate value of Δy as x varies from 1 to 2, if f(x) = 1.4 - 0.5x?

Learn how to approximate the change in y (Δy) using the derivative of a linear function f(x) = 1.4 - 0.5x. This problem demonstrates how small changes in x (Δx) affect y and uses key concepts of algebra and calculus, suitable for Grades 10-12.

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