Difference between revisions of "8A F11 Q12"

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<math>\begin{array}{rcl}
 
<math>\begin{array}{rcl}
 
\frac{2(3x + 1) -2(3(x + h) + 1)}{h(3(x + h) + 1)(3x + 1))} & = & \frac{6x + 2 - 6x -6h -2}{h(3(x + h) + 1)(3x + 1))}\\
 
\frac{2(3x + 1) -2(3(x + h) + 1)}{h(3(x + h) + 1)(3x + 1))} & = & \frac{6x + 2 - 6x -6h -2}{h(3(x + h) + 1)(3x + 1))}\\
& = & \frac{-6}{h(3(x + h) + 1)(3x + 1))}
+
& = & \frac{-6}{(3(x + h) + 1)(3x + 1))}
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
  
h(3(x + h) + 1)(3x + 1))}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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! Final Answer:
 +
|-
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|<math>\frac{-6}{(3(x + h) + 1)(3x + 1))}</math>
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|}
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[[8AF11Final|<u>'''Return to Sample Exam</u>''']]

Latest revision as of 22:57, 13 April 2015

Question: Find and simplify the difference quotient Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{f(x+h)-f(x)}{h}} for f(x) = Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2}{3x+1}}

Foundations
1) f(x + h) = ?
2) How do you eliminate the 'h' in the denominator?
Answer:
1) Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x + h) = \frac{2}{3(x + h) + 1}} the difference quotient is a difference of fractions divided by h.
2) The numerator is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2}{3(x + h) + 1} - \frac{2}{3x + 1}} so the first step is to simplify this expression. This then allows us to eliminate the 'h' in the denominator.

Solution:

Step 1:
The difference quotient that we want to simplify is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{f(x + h) - f(x)}{h} = \left(\frac{2}{3(x + h) + 1} - \frac{2}{3x + 1}\right) \div h}
Step 2:
Now we simplify the numerator:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \frac{f(x + h) - f(x)}{h} &=& \left(\frac{2}{3(x + h) + 1} - \frac{2}{3x + 1}\right) \div h\\ &=& \frac{2(3x + 1) -2(3(x + h) + 1)}{h(3(x + h) + 1)(3x + 1))} \end{array}}

Arithmetic:
Now we simplify the numerator:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \frac{2(3x + 1) -2(3(x + h) + 1)}{h(3(x + h) + 1)(3x + 1))} & = & \frac{6x + 2 - 6x -6h -2}{h(3(x + h) + 1)(3x + 1))}\\ & = & \frac{-6}{(3(x + h) + 1)(3x + 1))} \end{array}}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{-6}{(3(x + h) + 1)(3x + 1))}}

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