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The problem asks you to find the derivative of the function \( f(x) = 5x^3 + 7 \) using the definition of the derivative, and then evaluate the derivative at the points \( x = -3 \), \( x = 0 \), and \( x = 4 \) if the derivative exists.

### 1. Finding \( f'(x) \) using the definition of the derivative:

The definition of the derivative is given by:

\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]

Substitute \( f(x) = 5x^3 + 7 \) into the definition:

\[
f'(x) = \lim_{h \to 0} \frac{(5(x+h)^3 + 7) - (5x^3 + 7)}{h}
\]

### 2. Expanding \( (x+h)^3 \):

\[
(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3
\]

So,

\[
5(x+h)^3 = 5(x^3 + 3x^2h + 3xh^2 + h^3) = 5x^3 + 15x^2h + 15xh^2 + 5h^3
\]

Now, substitute back into the limit:

\[
f'(x) = \lim_{h \to 0} \frac{[5x^3 + 15x^2h + 15xh^2 + 5h^3 + 7] - [5x^3 + 7]}{h}
\]

Simplify the expression:

\[
f'(x) = \lim_{h \to 0} \frac{15x^2h + 15xh^2 + 5h^3}{h}
\]

Factor \( h \) out of the numerator:

\[
f'(x) = \lim_{h \to 0} \frac{h(15x^2 + 15xh + 5h^2)}{h}
\]

Cancel \( h \) in the numerator and denominator:

\[
f'(x) = \lim_{h \to 0} (15x^2 + 15xh + 5h^2)
\]

Take the limit as \( h \) approaches 0:

\[
f'(x) = 15x^2
\]

So the derivative of the function is \( f'(x) = 15x^2 \).

### 3. Evaluating \( f'(x) \) at specific points:

- \( f'(-3) = 15(-3)^2 = 15 \times 9 = 135 \)
- \( f'(0) = 15(0)^2 = 0 \)
- \( f'(4) = 15(4)^2 = 15 \times 16 = 240 \)

### Summary:

- The derivative of \( f(x) = 5x^3 + 7 \) is \( f'(x) = 15x^2 \).
- \( f'(-3) = 135 \)
- \( f'(0) = 0 \)
- \( f'(4) = 240 \)

Would you like further details or have any questions about this process? 

Here are some related questions:

1. What is the importance of the definition of the derivative in calculus?
2. How does the power rule for differentiation compare to the definition of the derivative?
3. What happens if we evaluate the derivative at points where the function is not continuous?
4. How can the concept of limits help us understand the derivative more intuitively?
5. How would the derivative change if the function were \( f(x) = 5x^3 + 7x + 1 \)?

**Tip:** Remember that the derivative can be interpreted as the slope of the tangent line to the curve at any given point.

Using the definition of the derivative, find f'(x). Then find f'(-3), f'(0), and f'(4) when the derivative exists, given f(x) = 5x^3 + 7.

Learn how to find the derivative of f(x) = 5x^3 + 7 using the definition of the derivative and evaluate the derivative at specific points x = -3, 0, and 4. Suitable for calculus students in Grades 11-12.

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