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Calculus
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Use the limit definition of the derivative to find \(\displaystyle f'(x)\) if \(\displaystyle f(x)= - 2 x^{2} + 9 x - 1\).

Using the difference quotient \(\displaystyle \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\) gives \(\displaystyle \lim_{h\to 0}\frac{(9 h + 9 x - 2 \left(h + x\right)^{2} - 1)-(- 2 x^{2} + 9 x - 1)}{h}\) Expanding and collecting like terms gives \(\displaystyle \lim_{h \to 0}\frac{- 2 h^{2} - 4 h x + 9 h}{h}\) Factoring out \(\displaystyle h\) and reducing gives \(\displaystyle \lim_{h\to 0}- 2 h - 4 x + 9\). Evaluating the limit gives \(\displaystyle f'(x)=- 4 x + 9\)

Download \(\LaTeX\) 

\begin{question}Use the limit definition of the derivative to find $f'(x)$ if $f(x)= - 2 x^{2} + 9 x - 1$. 
    \soln{9cm}{Using the difference quotient $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$ gives $\lim_{h\to 0}\frac{(9 h + 9 x - 2 \left(h + x\right)^{2} - 1)-(- 2 x^{2} + 9 x - 1)}{h}$ Expanding and collecting like terms gives $\lim_{h \to 0}\frac{- 2 h^{2} - 4 h x + 9 h}{h}$ Factoring out $h$ and reducing gives $\lim_{h\to 0}- 2 h - 4 x + 9$. Evaluating the limit gives $f'(x)=- 4 x + 9$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<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)= - 2 x^{2} + 9 x - 1 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3D%20-%202%20x%5E%7B2%7D%20%2B%209%20x%20-%201%20" alt="LaTeX:  \displaystyle f(x)= - 2 x^{2} + 9 x - 1 " data-equation-content=" \displaystyle f(x)= - 2 x^{2} + 9 x - 1 " /> . </p> </p>
HTML for Canvas
<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{(9 h + 9 x - 2 \left(h + x\right)^{2} - 1)-(- 2 x^{2} + 9 x - 1)}{h} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto%200%7D%5Cfrac%7B%289%20h%20%2B%209%20x%20-%202%20%5Cleft%28h%20%2B%20x%5Cright%29%5E%7B2%7D%20-%201%29-%28-%202%20x%5E%7B2%7D%20%2B%209%20x%20-%201%29%7D%7Bh%7D%20" alt="LaTeX:  \displaystyle \lim_{h\to 0}\frac{(9 h + 9 x - 2 \left(h + x\right)^{2} - 1)-(- 2 x^{2} + 9 x - 1)}{h} " data-equation-content=" \displaystyle \lim_{h\to 0}\frac{(9 h + 9 x - 2 \left(h + x\right)^{2} - 1)-(- 2 x^{2} + 9 x - 1)}{h} " />  Expanding and collecting like terms gives  <img class="equation_image" title=" \displaystyle \lim_{h \to 0}\frac{- 2 h^{2} - 4 h x + 9 h}{h} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%20%5Cto%200%7D%5Cfrac%7B-%202%20h%5E%7B2%7D%20-%204%20h%20x%20%2B%209%20h%7D%7Bh%7D%20" alt="LaTeX:  \displaystyle \lim_{h \to 0}\frac{- 2 h^{2} - 4 h x + 9 h}{h} " data-equation-content=" \displaystyle \lim_{h \to 0}\frac{- 2 h^{2} - 4 h x + 9 h}{h} " />  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}- 2 h - 4 x + 9 " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bh%5Cto%200%7D-%202%20h%20-%204%20x%20%2B%209%20" alt="LaTeX:  \displaystyle \lim_{h\to 0}- 2 h - 4 x + 9 " data-equation-content=" \displaystyle \lim_{h\to 0}- 2 h - 4 x + 9 " /> . Evaluating the limit gives  <img class="equation_image" title=" \displaystyle f'(x)=- 4 x + 9 " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%3D-%204%20x%20%2B%209%20" alt="LaTeX:  \displaystyle f'(x)=- 4 x + 9 " data-equation-content=" \displaystyle f'(x)=- 4 x + 9 " /> </p> </p>