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### Problem 5:
"An object is sent through the air. Its height is modelled by the function \( h(x) = -5x^2 + 3x + 65 \), where \( h(x) \) is the height of the object in meters, and \( x \) is the time in seconds. Estimate the instantaneous rate of change in the object's height at \( x = 3 \) seconds."

To solve this, we'll compute the derivative of the height function, \( h(x) \), to get the instantaneous rate of change, and then substitute \( x = 3 \).

### Solution:
1. The height function is \( h(x) = -5x^2 + 3x + 65 \).
2. Differentiate \( h(x) \) to find the rate of change function, \( h'(x) \):
   \[
   h'(x) = \frac{d}{dx}[-5x^2 + 3x + 65] = -10x + 3
   \]
3. To estimate the instantaneous rate of change at \( x = 3 \) seconds, substitute \( x = 3 \) into \( h'(x) \):
   \[
   h'(3) = -10(3) + 3 = -30 + 3 = -27
   \]

Thus, the instantaneous rate of change in the object's height at \( x = 3 \) seconds is \(-27\) meters per second. This means the object is falling at a rate of 27 m/s at that moment.

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Here are 5 related questions to expand your understanding:

1. What is the geometric interpretation of the derivative \( h'(x) \) in the context of this problem?
2. How would the object's motion change if the coefficient of \( x^2 \) in \( h(x) \) were positive?
3. What is the meaning of the constant term 65 in the function \( h(x) \)?
4. How does the sign of \( h'(x) \) indicate whether the object is rising or falling at a given point?
5. How can you estimate instantaneous rates of change if a function is given in a table rather than an equation?

### Tip:
When estimating instantaneous rates of change from a function, always remember that the derivative gives the slope of the tangent line at any point, which corresponds to the rate of change at that exact moment.

An object is sent through the air. Its height is modelled by the function h(x) = -5x^2 + 3x + 65, where h(x) is the height of the object in metres, and x is the time in seconds. Estimate the instantaneous rate of change in the object's height at 3 seconds.

Solve the problem of estimating the instantaneous rate of change in an object's height modelled by the quadratic function h(x) = -5x^2 + 3x + 65 at 3 seconds. Learn key calculus concepts and derivative rules for quadratic functions.

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