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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).

User Moudiz
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1 Answer

4 votes

Answer:

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
\(x\).

User Chris Banes
by
8.4k points