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Find [tex]\(\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\)[/tex] for the given function and value of [tex]\(x\)[/tex].

Given: [tex]\( f(x) = -2x - 1 \)[/tex], [tex]\( x = 6 \)[/tex]

The [tex]\(\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\)[/tex] for [tex]\( f(x) = -2x - 1 \)[/tex], [tex]\( x = 6 \)[/tex] is [tex]\(\boxed{\phantom{0}}\)[/tex].

(Type an integer or a simplified fraction.)

Sagot :

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].