Answer:
f ′ (x)=3.99x 2 +7.98x+45.06
Explanation:
Let's break down the process of finding the derivative of the function
in simpler terms.
We want to find the derivative
, which tells us how the function's value changes as
changes. It's like finding the slope of the function at any point.
1. Start with the function

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
, which we'll call
. The formula is:
![\[f'(x) = \lim_(h \to 0) (f(x + h) - f(x))/(h)\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/2g4vw7sva28unvr9yrnj5l0v1i4vf65k6b.png)
In plain terms, it means: "Find the difference in
when
changes by a very small amount
and divide it by that small amount
"
3. Plug our function \(f(x)\) into the formula:
![\[f'(x) = \lim_(h \to 0) (3.99(x + h)^2 + 45.06(x + h) + 251.7 - (3.99x^2 + 45.06x + 251.7))/(h)\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/k4ophxujgcbezxhcplh7wellrxrjkf4lej.png)
This represents the change in

4. Simplify the expression inside the limit:
![\[f'(x) = \lim_(h \to 0) (3.99x^2 + 7.98xh + 3.99h^2 + 45.06h)/(h)\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/2j6prbjlvtunpeqpxq5vuxrmnko4szh0iu.png)
This breaks down how
changes when
changes by

5. Cancel out \(h\) in the numerator and denominator:
![\[f'(x) = \lim_(h \to 0) (3.99x^2 + 7.98x + 3.99h + 45.06)\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/hjdx5gptw5ww32em3fn59pp18hsh1ubzw7.png)
This tells us the rate of change of
as
gets smaller and smaller.
6. Finally, as
approaches 0 (gets extremely small), the term
and
become very close to 0. So, we're left with:
![\[f'(x) = 3.99x^2 + 7.98x + 45.06\]](https://img.qammunity.org/2024/formulas/mathematics/high-school/w5gdgu9iza5emnj26h4ovkuw1pbzrq1nti.png)
This is the derivative
of the original function. It represents how the original function's value changes with respect to
