To find [tex]\(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)[/tex] for the function [tex]\(f(x) = -2x - 1\)[/tex] at [tex]\(x = 6\)[/tex], we follow these steps:
1. Determine [tex]\(f(x+h)\)[/tex]:
Given [tex]\(f(x) = -2x - 1\)[/tex], we substitute [tex]\(x + h\)[/tex] into the function:
[tex]\[
f(x+h) = -2(x+h) - 1
\][/tex]
Simplifying this, we get:
[tex]\[
f(x+h) = -2x - 2h - 1
\][/tex]
2. Find the difference [tex]\(f(x+h) - f(x)\)[/tex]:
We already have [tex]\(f(x) = -2x - 1\)[/tex] and [tex]\(f(x+h) = -2x - 2h - 1\)[/tex]. Now, we calculate:
[tex]\[
f(x+h) - f(x) = (-2x - 2h - 1) - (-2x - 1)
\][/tex]
Simplifying this, we get:
[tex]\[
f(x+h) - f(x) = -2x - 2h - 1 + 2x + 1
\][/tex]
[tex]\[
f(x+h) - f(x) = -2h
\][/tex]
3. Form the difference quotient [tex]\(\frac{f(x+h) - f(x)}{h}\)[/tex]:
We now put our result into the difference quotient:
[tex]\[
\frac{f(x+h) - f(x)}{h} = \frac{-2h}{h}
\][/tex]
Simplifying this, we get:
[tex]\[
\frac{f(x+h) - f(x)}{h} = -2
\][/tex]
4. Take the limit as [tex]\(h \to 0\)[/tex]:
Since the quotient [tex]\(-2\)[/tex] does not depend on [tex]\(h\)[/tex] (it is constant), the limit is:
[tex]\[
\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = -2
\][/tex]
Therefore, the limit [tex]\(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)[/tex] for the given function [tex]\(f(x) = -2x - 1\)[/tex] at [tex]\(x = 6\)[/tex] is [tex]\(-2\)[/tex].
