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Let's work through this step-by-step.

Given:
\[
f(x) = x^2 + 3x - 5
\]

### Part (a)
To estimate the instantaneous rate of change at \( x = 1 \), we need to find the derivative \( f'(x) \) and evaluate it at \( x = 1 \). However, since this part asks for an estimation, let's use the definition of the derivative:

\[
f'(1) \approx \frac{f(1+h) - f(1)}{h}
\]
where \( h \) is a small value. Let's use a small value for \( h \) (like \( h = 0.01 \)) for the estimation:
- \( f(1) = 1^2 + 3 \cdot 1 - 5 = -1 \)
- \( f(1 + 0.01) = (1 + 0.01)^2 + 3 \cdot (1 + 0.01) - 5 \)

Calculate the values and then approximate the rate of change.

### Part (b)
We are asked to simplify the expression:
\[
\frac{f(x+h) - f(x)}{h}
\]
Substitute \( f(x) = x^2 + 3x - 5 \):

1. \( f(x+h) = (x+h)^2 + 3(x+h) - 5 \)
2. Expand \( f(x+h) \):
   \[
   f(x+h) = x^2 + 2xh + h^2 + 3x + 3h - 5
   \]
3. Compute \( f(x+h) - f(x) \):
   \[
   f(x+h) - f(x) = (x^2 + 2xh + h^2 + 3x + 3h - 5) - (x^2 + 3x - 5)
   \]
4. Simplify this expression by canceling terms:
   \[
   f(x+h) - f(x) = 2xh + h^2 + 3h
   \]
5. Factor out \( h \):
   \[
   \frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 + 3h}{h} = 2x + h + 3
   \]

### Part (c)
Examine the expression \( 2x + h + 3 \) as \( h \to 0 \). As \( h \) approaches 0, the expression simplifies to:
\[
2x + 3
\]

### Part (d)
Using the result from part (c), we find that the expression for the instantaneous rate of change (the derivative) at any point \( x \) is:
\[
f'(x) = 2x + 3
\]
To verify this with the result from part (a) for \( x = 1 \):
\[
f'(1) = 2 \cdot 1 + 3 = 5
\]

Thus, the instantaneous rate of change at \( x = 1 \) is indeed 5, confirming our result from part (a).

---

Would you like any further details or have additional questions?

Here are 5 related questions to further expand on this topic:
1. What is the geometric interpretation of the instantaneous rate of change?
2. How would the rate of change change if the function were \( f(x) = x^2 + 3x - 5 + c \), where \( c \) is a constant?
3. How does the concept of a derivative connect to the slope of a tangent line?
4. What is the significance of the limit process in defining derivatives?
5. How would you find the second derivative of \( f(x) = x^2 + 3x - 5 \) and what does it represent?

**Tip:** When estimating the instantaneous rate of change, choosing a smaller \( h \)-value makes the estimation closer to the actual derivative value.

Consider the function f(x) = x^2 + 3x - 5. 
a) Estimate the instantaneous rate of change at x = 1. 
b) Simplify the expression (f(x + h) - f(x)) / h. 
c) Examine the expression in part b) and discuss what happens as h becomes very close to 0. 
d) Use your result from part c) to come up with an expression for the instantaneous rate of change at the point x, and check your result from part a) using the expression.

Learn how to estimate the instantaneous rate of change at a specific point, simplify derivative expressions, and analyze limits for the function f(x) = x^2 + 3x - 5. This solution walks through key calculus concepts, including using the limit definition of a derivative. Suitable for students in Grades 11-12.

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