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To find the gradient of the given function \( y = 5 + 8x - \frac{4}{3}x^3 \) at \( x = -1 \), we need to differentiate the function with respect to \( x \) and then evaluate the derivative at \( x = -1 \).

### Step 1: Differentiate the function

The function is \( y = 5 + 8x - \frac{4}{3}x^3 \). We will differentiate each term:

- The derivative of \( 5 \) is \( 0 \).
- The derivative of \( 8x \) is \( 8 \).
- The derivative of \( -\frac{4}{3}x^3 \) is \( -\frac{4}{3} \cdot 3x^2 = -4x^2 \).

So, the derivative (gradient function) is:
\[
\frac{dy}{dx} = 8 - 4x^2
\]

### Step 2: Evaluate the derivative at \( x = -1 \)

Now, substitute \( x = -1 \) into the derivative:
\[
\frac{dy}{dx} = 8 - 4(-1)^2 = 8 - 4(1) = 8 - 4 = 4
\]

### Final Answer:

The gradient of the function at \( x = -1 \) is \( 4 \).

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Would you like more details on the differentiation steps or have any questions? Here are some related questions:

1. How do we differentiate more complex functions like trigonometric or logarithmic expressions?
2. What does the gradient of a function tell us about its behavior at a specific point?
3. How can the power rule for differentiation be extended to fractional powers of \( x \)?
4. How does differentiation work when we deal with products or quotients of functions?
5. What is the significance of second derivatives, and how do they relate to concavity?

**Tip**: To find the gradient of a curve at any point, always remember to differentiate first, and then substitute the specific \( x \)-value into the derivative!

Find the gradient of y = 5 + 8x - (4/3)x^3 at x = -1.

Learn how to find the gradient of the function y = 5 + 8x - (4/3)x^3 at x = -1 by differentiating the function and substituting the value of x. This calculus problem uses the power rule for differentiation and is suitable for high school students in grades 11-12.

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