To address your request, let's first find the rate of change of the function $f(t) = 150 \cdot 1.126^{10}$, and then proceed to find the linearization of $f(t)$.

1. Finding the rate of change:

The function given is a constant function, as $1.126^{10}$ is a number independent of $t$. Let's calculate it:

$f(t) = 150 \cdot 1.126^{10} \approx 150 \cdot 3.239 = 485.85$

Since there is no variable $t$ involved, the rate of change (which refers to the derivative) is 0. Therefore, the derivative is:

$f'(t) = 0$

2. Linearization of the function:

For the linearization, we usually find the tangent line to the function at a point $t = a$, but since $f(t)$ is constant (no dependence on $t$), its linearization is simply the constant value of the function:

$L(t) = f(a) + f'(a)(t - a) = 485.85 + 0 \cdot (t - a) = 485.85$

Thus, the linearization of $f(t)$ is $L(t) = 485.85$, which is just a horizontal line.

Would you like more details or further clarification on this? Here are five related questions:

1. What is the interpretation of the linearization for a constant function?
2. How would you linearize a function that actually depends on $t$?
3. How does the rate of change (derivative) behave for functions of different types (e.g., polynomials, exponentials)?
4. What is the general formula for the linear approximation of a differentiable function?
5. How do second derivatives provide insights into the curvature or concavity of a function?

Tip: For a constant function, the derivative is always zero, and its linear approximation remains the constant itself, since there is no variation in the slope.