The image shows a curve representing a function $g(x)$ and a tangent line at point $B$ where $B(1, 4)$. A nearby point on the curve is given as $(0.95, 4.03)$.

We are asked to fill in the blanks:

Part (a):

$g(1) = 4$ is already provided, indicating the function value at point $B$.

Part (b):

We need to calculate $g'(1)$, which is the derivative of $g$ at $x = 1$, representing the slope of the tangent line at point $B$.

To find $g'(1)$, we can approximate the derivative using the given points:

$g'(1) \approx \frac{g(0.95) - g(1)}{0.95 - 1}$

Substituting the values:

$g'(1) \approx \frac{4.03 - 4}{0.95 - 1} = \frac{0.03}{-0.05} = -0.6$

Thus, $g'(1) = -0.6$.

Final answers:

• (a) $g(1) = 4$
• (b) $g'(1) = -0.6$

Do you need further details or clarifications?
Here are some related questions to explore:

1. What is the geometric meaning of the derivative at a point?
2. How do you interpret the sign of the derivative in terms of the curve's behavior?
3. What other methods can be used to approximate derivatives?
4. How would the approximation change if a different nearby point were used?
5. What is the significance of the tangent line in understanding the function locally?

Tip: The slope of the tangent line gives key information about the rate of change of the function at a specific point.