Let's solve the problems in the image using linear approximation (also known as tangent line approximation).

Problem 1:

We are given a function $f(x)$ and its tangent line at the point $(-6, 3)$ with the equation: $y = 3 - 4(x + 6)$ This can be written as: $y = 3 - 4x - 24 \quad \Rightarrow \quad y = -4x - 21$ This tangent line gives the linear approximation of $f(x)$ around $x = -6$. We are asked to approximate $f(-4)$.

Substitute $x = -4$ into the tangent line equation: $f(-4) \approx -4(-4) - 21 = 16 - 21 = -5$

So, the approximate value of $f(-4)$ is: $f(-4) \approx -5$

Problem 2:

We are given that $g(-6) = 8$ and $g'(-6) = -6$. The formula for the linear approximation of $g(x)$ at $x = -6$ is: $g(x) \approx g(-6) + g'(-6)(x - (-6)) = 8 + (-6)(x + 6)$ Simplify this expression: $g(x) \approx 8 - 6(x + 6) = 8 - 6x - 36 = -6x - 28$

So the linear approximation formula is: $g(x) \approx -6x - 28$

Problem 3:

We are given that $h(5) = -2$ and $h'(5) = 2$. We need to use linear approximation to approximate $h(3.5)$. The formula for the linear approximation of $h(x)$ at $x = 5$ is: $h(x) \approx h(5) + h'(5)(x - 5) = -2 + 2(x - 5)$ Substitute $x = 3.5$: $h(3.5) \approx -2 + 2(3.5 - 5) = -2 + 2(-1.5) = -2 - 3 = -5$

So, the approximate value of $h(3.5)$ is: $h(3.5) \approx -5$

Do you want further details or have any questions on the process? Here are five related questions:

1. How do we derive the equation of a tangent line for a given function?
2. Why does linear approximation work well for values near the point of tangency?
3. How can we improve the accuracy of the approximation for values farther from the base point?
4. What is the difference between a tangent line approximation and a secant line approximation?
5. How is linear approximation related to the concept of differentials in calculus?

Tip: Linear approximation works best when the point you are approximating is close to the point where you know the exact value. The further away you get, the less accurate the approximation becomes.