Given $f(x)=\frac{3}{x+2}$ , find $f^{\prime}(1)$ using the definition of a derivative.

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Algebra 2

Evaluate expression when two complex numbers are given

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Q. Given

f(x)=\frac{3}{x+2}

, find

f^{\prime}(1)

using the definition of a derivative.

Given Function: We are given the function $f(x) = \frac{3}{(x+2)}$ and we need to find its derivative at $x = 1$ using the definition of a derivative. The definition of a derivative is: $\newline$ $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ $\newline$ Let's apply this definition to our function.
Substitute $x+h$ : First, we need to find the expression for $f(x+h)$ . This means we will substitute $(x+h)$ for $x$ in the function $f(x)$ : $f(x+h) = \frac{3}{((x+h)+2)} = \frac{3}{x+h+2}$
Plug into Derivative Definition: Now, we will plug $f(x+h)$ and $f(x)$ into the definition of the derivative: $f'(x) = \lim_{h \to 0} \left[\frac{3}{(x+h+2)} - \frac{3}{(x+2)}\right] / h$
Find Common Denominator: Next, we need to find a common denominator for the terms in the numerator to combine them: $\newline$ The common denominator is $(x+h+2)(x+2)$ , so we rewrite the numerator: $\newline$ $f'(x) = \lim_{h \to 0} \left[\frac{3(x+2) - 3(x+h+2)}{(x+h+2)(x+2)}\right] / h$
Simplify Numerator: Now, we simplify the numerator: $\newline$ $3(x+2) - 3(x+h+2) = 3x + 6 - 3x - 3h - 6 = -3h$ $\newline$ So the expression becomes: $\newline$ $f'(x) = \lim_{h \to 0} \left[\frac{-3h}{((x+h+2)(x+2) \cdot h)}\right]$
Cancel out h: We can now cancel out the h in the numerator and denominator: $f'(x) = \lim_{h \to 0} \left[-\frac{3}{((x+h+2)(x+2))}\right]$
Take the Limit: Now we can take the limit as $h$ approaches $0$ : $f'(x) = \frac{-3}{((x+2)(x+2))}$
Substitute $x=1$ : Finally, we substitute $x = 1$ into the derivative to find $f'(1)$ : $f'(1) = -\frac{3}{((1+2)(1+2))} = -\frac{3}{(3\times3)} = -\frac{3}{9} = -\frac{1}{3}$