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Questions: Algebra BusinessCalculus
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Use the limit definition of the derivative to find \(\displaystyle f'(x)\) if \(\displaystyle f(x)= \frac{3}{x - 5}\).
Using the difference quotient \(\displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\) gives \(\displaystyle \lim_{h\to 0}\frac{\frac{3}{h + x - 5}-\frac{3}{x - 5}}{h}\) Using the least common denominator, \(\displaystyle (h + x - 5)(x - 5) \), to clear the fractions gives \(\displaystyle \frac{3(x - 5)-3(h + x - 5)}{h(x - 5)(h + x - 5)}\). Expanding gives \(\displaystyle \frac{- 3 h}{h \left(x - 5\right) \left(h + x - 5\right)}\). Factoring out \(\displaystyle h\) and reducing gives \(\displaystyle \lim_{h\to 0}- \frac{3}{\left(x - 5\right) \left(h + x - 5\right)}\). Evaluating the limit gives \(\displaystyle f'(x)=- \frac{3}{\left(x - 5\right)^{2}}\)
\begin{question}Use the limit definition of the derivative to find $f'(x)$ if $f(x)= \frac{3}{x - 5}$.
\soln{9cm}{Using the difference quotient $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$ gives $\lim_{h\to 0}\frac{\frac{3}{h + x - 5}-\frac{3}{x - 5}}{h}$ Using the least common denominator, $(h + x - 5)(x - 5) $, to clear the fractions gives $\frac{3(x - 5)-3(h + x - 5)}{h(x - 5)(h + x - 5)}$. Expanding gives $\frac{- 3 h}{h \left(x - 5\right) \left(h + x - 5\right)}$. Factoring out $h$ and reducing gives $\lim_{h\to 0}- \frac{3}{\left(x - 5\right) \left(h + x - 5\right)}$. Evaluating the limit gives $f'(x)=- \frac{3}{\left(x - 5\right)^{2}}$}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Use the limit definition of the derivative to find <img class="equation_image" title=" \displaystyle f'(x) " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20" alt="LaTeX: \displaystyle f'(x) " data-equation-content=" \displaystyle f'(x) " /> if <img class="equation_image" title=" \displaystyle f(x)= \frac{3}{x - 5} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3D%20%5Cfrac%7B3%7D%7Bx%20-%205%7D%20" alt="LaTeX: \displaystyle f(x)= \frac{3}{x - 5} " data-equation-content=" \displaystyle f(x)= \frac{3}{x - 5} " /> . </p> </p><p> <p>Using the difference quotient <img class="equation_image" title=" \displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%20%5Cto%200%7D%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%20" alt="LaTeX: \displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} " data-equation-content=" \displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} " /> gives <img class="equation_image" title=" \displaystyle \lim_{h\to 0}\frac{\frac{3}{h + x - 5}-\frac{3}{x - 5}}{h} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto%200%7D%5Cfrac%7B%5Cfrac%7B3%7D%7Bh%20%2B%20x%20-%205%7D-%5Cfrac%7B3%7D%7Bx%20-%205%7D%7D%7Bh%7D%20" alt="LaTeX: \displaystyle \lim_{h\to 0}\frac{\frac{3}{h + x - 5}-\frac{3}{x - 5}}{h} " data-equation-content=" \displaystyle \lim_{h\to 0}\frac{\frac{3}{h + x - 5}-\frac{3}{x - 5}}{h} " /> Using the least common denominator, <img class="equation_image" title=" \displaystyle (h + x - 5)(x - 5) " src="/equation_images/%20%5Cdisplaystyle%20%28h%20%2B%20x%20-%205%29%28x%20-%205%29%20%20" alt="LaTeX: \displaystyle (h + x - 5)(x - 5) " data-equation-content=" \displaystyle (h + x - 5)(x - 5) " /> , to clear the fractions gives <img class="equation_image" title=" \displaystyle \frac{3(x - 5)-3(h + x - 5)}{h(x - 5)(h + x - 5)} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B3%28x%20-%205%29-3%28h%20%2B%20x%20-%205%29%7D%7Bh%28x%20-%205%29%28h%20%2B%20x%20-%205%29%7D%20" alt="LaTeX: \displaystyle \frac{3(x - 5)-3(h + x - 5)}{h(x - 5)(h + x - 5)} " data-equation-content=" \displaystyle \frac{3(x - 5)-3(h + x - 5)}{h(x - 5)(h + x - 5)} " /> . Expanding gives <img class="equation_image" title=" \displaystyle \frac{- 3 h}{h \left(x - 5\right) \left(h + x - 5\right)} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B-%203%20h%7D%7Bh%20%5Cleft%28x%20-%205%5Cright%29%20%5Cleft%28h%20%2B%20x%20-%205%5Cright%29%7D%20" alt="LaTeX: \displaystyle \frac{- 3 h}{h \left(x - 5\right) \left(h + x - 5\right)} " data-equation-content=" \displaystyle \frac{- 3 h}{h \left(x - 5\right) \left(h + x - 5\right)} " /> . Factoring out <img class="equation_image" title=" \displaystyle h " src="/equation_images/%20%5Cdisplaystyle%20h%20" alt="LaTeX: \displaystyle h " data-equation-content=" \displaystyle h " /> and reducing gives <img class="equation_image" title=" \displaystyle \lim_{h\to 0}- \frac{3}{\left(x - 5\right) \left(h + x - 5\right)} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto%200%7D-%20%5Cfrac%7B3%7D%7B%5Cleft%28x%20-%205%5Cright%29%20%5Cleft%28h%20%2B%20x%20-%205%5Cright%29%7D%20" alt="LaTeX: \displaystyle \lim_{h\to 0}- \frac{3}{\left(x - 5\right) \left(h + x - 5\right)} " data-equation-content=" \displaystyle \lim_{h\to 0}- \frac{3}{\left(x - 5\right) \left(h + x - 5\right)} " /> . Evaluating the limit gives <img class="equation_image" title=" \displaystyle f'(x)=- \frac{3}{\left(x - 5\right)^{2}} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%3D-%20%5Cfrac%7B3%7D%7B%5Cleft%28x%20-%205%5Cright%29%5E%7B2%7D%7D%20" alt="LaTeX: \displaystyle f'(x)=- \frac{3}{\left(x - 5\right)^{2}} " data-equation-content=" \displaystyle f'(x)=- \frac{3}{\left(x - 5\right)^{2}} " /> </p> </p>