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Given
f(x)=2x-3, find
f^(')(-2) using the definition of a derivative.

Given $f(x)=2 x-3$ , find $f^{\prime}(-2)$ using the definition of a derivative.

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Find derivatives using logarithmic differentiation

Full solution

Q. Given

f(x)=2 x-3

, find

f^{\prime}(-2)

using the definition of a derivative.

Define derivative limit: To find $f'(-2)$ , we need to use the definition of the derivative, which is the limit of the average rate of change as the change in $x$ approaches $0$ .
Substitute values: The definition of the derivative is $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ .
Calculate $f(-2+h)$ : Substitute $x$ with $-2$ and $f(x)$ with $2x-3$ into the definition: $f^{\prime}(-2) = \lim_{h\to 0} \frac{f(-2+h) - f(-2)}{h}$ .
Calculate $f(-2)$ : Calculate $f(-2+h)$ : $f(-2+h) = 2(-2+h)-3 = -4+2h-3$ .
Substitute into limit: Calculate $f(-2)$ : $f(-2) = 2(-2)-3 = -4-3 = -7$ .
Simplify expression: Substitute $f(-2+h)$ and $f(-2)$ into the limit: $f^{\prime}(-2) = \lim_{h\to 0} \frac{(-4+2h-3) - (-7)}{h}$ .
Further simplify: Simplify the expression inside the limit: $f^{\prime}(-2) = \lim_{h \to 0} \frac{[2h-4+3+7]}{h}$ .
Divide by h: Further simplify the expression: $f'(-2) = \lim_{h\to 0} \frac{2h+6}{h}$ .
Identify mistake: Divide each term by $h$ : $f'(-2) = \lim_{h \to 0} [2 + \frac{6}{h}]$ .
Identify mistake: Divide each term by $h$ : $f'(\$-2$ ) = \lim_{h\to $0$ } [ $2$ + \frac{ $6$ }{h}]\).As $h$ approaches $0$ , the term $\frac{6}{h}$ approaches infinity, which is incorrect since we expect a finite number for the derivative. There's a mistake in the previous step.

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