Let ( f(x)=frac{x-3}{2 x+1} ). For this problem, use the following definition of the derivative, ( f^{prime}(a)=lim _{x ightarrow a} frac{f(x)-f(a)}{x-a} ). a. Set up the difference quotient (do not simplify), ( lim _{x ightarrow a} ) b. The numerator of the difference quotient when simplified into a single fraction becomes c. Before taking the limit, and after simplifying, the difference quotient expression becomes ( lim _{x ightarrow a} ) d. Taking the limit, we have that ( f^{prime}(a)= )
Added by Cavaughn J.
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First, we need to set up the difference quotient. We have: \(f(x) = \frac{x-3}{2x+1}\) and \(f(a) = \frac{a-3}{2a+1}\) So, the difference quotient is: \(\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a} = \lim_{x \rightarrow a} \frac{\frac{x-3}{2x+1} - Show more…
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