To find the instantaneous rate of change at any point [tex]\(x\)[/tex] for the function [tex]\(f(x) = 2x^2 - x\)[/tex] using the definition of the derivative, let's go through the following steps:
1. Expression for [tex]\(f(x + h)\)[/tex]:
We need to find the value of the function when [tex]\(x\)[/tex] is incremented by a small value [tex]\(h\)[/tex]. Specifically, we find [tex]\(f(x + h)\)[/tex].
[tex]\[
f(x + h) = 2(x + h)^2 - (x + h)
\][/tex]
Expanding this expression, we have:
[tex]\[
f(x + h) = 2(x^2 + 2xh + h^2) - x - h
\][/tex]
Simplifying this, we get:
[tex]\[
f(x + h) = 2x^2 + 4xh + 2h^2 - x - h
\][/tex]
2. Difference [tex]\(f(x + h) - f(x)\)[/tex]:
Now, we subtract [tex]\(f(x)\)[/tex] from [tex]\(f(x + h)\)[/tex]:
[tex]\[
f(x + h) - f(x) = (2x^2 + 4xh + 2h^2 - x - h) - (2x^2 - x)
\][/tex]
Simplifying this, we get:
[tex]\[
f(x + h) - f(x) = 2x^2 + 4xh + 2h^2 - x - h - 2x^2 + x
\][/tex]
[tex]\[
f(x + h) - f(x) = 4xh + 2h^2 - h
\][/tex]
So, the result is:
[tex]\[
f(x + h) - f(x) = 4xh + 2h^2 - h
\][/tex]
3. Quotient [tex]\(\frac{f(x + h) - f(x)}{h}\)[/tex]:
We now divide the difference by [tex]\(h\)[/tex]:
[tex]\[
\frac{f(x + h) - f(x)}{h} = \frac{4xh + 2h^2 - h}{h}
\][/tex]
Simplifying this, we get:
[tex]\[
\frac{f(x + h) - f(x)}{h} = 4x + 2h - 1
\][/tex]
4. Limit as [tex]\(h\)[/tex] approaches 0:
Finally, we take the limit of the quotient as [tex]\(h \rightarrow 0\)[/tex]:
[tex]\[
\lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} (4x + 2h - 1)
\][/tex]
Since the term [tex]\(2h\)[/tex] goes to 0 as [tex]\(h\)[/tex] approaches 0, the limit simplifies to:
[tex]\[
\lim_{h \rightarrow 0} (4x + 2h - 1) = 4x - 1
\][/tex]
So, the instantaneous rate of change, or the derivative of the function [tex]\(f(x) = 2x^2 - x\)[/tex], at any point [tex]\(x\)[/tex] is:
[tex]\[
f'(x) = 4x - 1
\][/tex]
