Question

The yearly enrollment at a certain university can be modeled by the function. f(x) = 3.99x² + 45.06x + 251.7, 1 ≤ x ≤ 10, where x is the number of year since 1992. Use the definition of the derivative to determine f'(x).

Answer 1

f ′ (x)=3.99x 2 +7.98x+45.06

Explanation:

Let's break down the process of finding the derivative of the function
`\(f(x) = 3.99x^2 + 45.06x + 251.7\)` in simpler terms.

We want to find the derivative
`\(f'(x)\)`, which tells us how the function's value changes as
`\(x\)`changes. It's like finding the slope of the function at any point.

1. Start with the function
`\(f(x) = 3.99x^2 + 45.06x + 251.7\).`

2. We'll use a formula to calculate the derivative. This formula looks at how the function changes when we make a tiny change in
`\(x\)`, which we'll call
`\(h\)`. The formula is:

`\[f'(x) = \lim_(h \to 0) (f(x + h) - f(x))/(h)\]`

In plain terms, it means: "Find the difference in
`\(f(x)\)`when
`\(x\)`changes by a very small amount
`\(h\),` and divide it by that small amount
`\(h\).`"

3. Plug our function \(f(x)\) into the formula:

`\[f'(x) = \lim_(h \to 0) (3.99(x + h)^2 + 45.06(x + h) + 251.7 - (3.99x^2 + 45.06x + 251.7))/(h)\]`

This represents the change in
`\(f(x)\) when \(x\) changes by \(h\).`

4. Simplify the expression inside the limit:

`\[f'(x) = \lim_(h \to 0) (3.99x^2 + 7.98xh + 3.99h^2 + 45.06h)/(h)\]`

This breaks down how
`\(f(x)\)` changes when
`\(x\)`changes by
`\(h\).`

5. Cancel out \(h\) in the numerator and denominator:

`\[f'(x) = \lim_(h \to 0) (3.99x^2 + 7.98x + 3.99h + 45.06)\]`

This tells us the rate of change of
`\(f(x)\)` as
`\(h\)` gets smaller and smaller.

6. Finally, as
`\(h\)` approaches 0 (gets extremely small), the term
`\(3.99h\)` and
`\(3.99h^2\)` become very close to 0. So, we're left with:

`\[f'(x) = 3.99x^2 + 7.98x + 45.06\]`

This is the derivative
`\(f'(x)\)` of the original function. It represents how the original function's value changes with respect to
`\(x\).`