The problem in the image has two parts involving a function $g(t) = (0.4)^t$. Let's go through both parts step by step.

Part (a)

You are asked to sketch a graph of $g(t) = (0.4)^t$ and determine whether $g'(2)$ is positive, negative, or zero.

Solution:

The function $g(t) = (0.4)^t$ is an exponential function with base less than 1, which means the graph is decreasing. The derivative of a function gives us the slope of the tangent line at any point, so to determine whether $g'(2)$ is positive, negative, or zero, we need to understand the behavior of the graph at $t = 2$.

Since $g(t) = (0.4)^t$ is a decreasing function, the slope at $t = 2$ is negative. Thus, we can conclude that:

• Answer: Negative

Part (b)

You are asked to use a small interval to estimate $g'(2)$ and round the answer to three decimal places.

Solution:

The derivative $g'(2)$ can be approximated using the definition of the derivative: $g'(2) \approx \frac{g(2 + h) - g(2)}{h}$ for a small value of $h$. Let's choose $h = 0.001$ and compute an estimate for $g'(2)$.

We know $g(t) = (0.4)^t$, so:

$g(2) = (0.4)^2 = 0.16$ $g(2 + h) = (0.4)^{2.001} \approx 0.1596 \quad \text{(using calculator approximation)}$

Now, using the values for $g(2)$ and $g(2 + h)$:

$g'(2) \approx \frac{0.1596 - 0.16}{0.001} = \frac{-0.0004}{0.001} = -0.4$

So, the estimated value of $g'(2)$ is approximately -0.400.

Summary of answers:

• (a) The derivative $g'(2)$ is Negative.
• (b) The estimate for $g'(2)$ is approximately -0.400.

Would you like more details on any of the steps?

Here are some related questions you could explore:

1. How is the derivative of exponential functions with base less than 1 generally computed?
2. Why do functions with bases less than 1 always have negative slopes?
3. How would the graph of $g(t) = (0.4)^t$ look for larger values of $t$?
4. What is the geometric interpretation of the derivative at a point?
5. How does changing the base in an exponential function affect the behavior of its graph?

Tip: For small changes in $t$, using the difference quotient formula is a great way to approximate the derivative if calculating it directly is complicated.