Difference between revisions of "8A F11 Q12"
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\end{array}</math> | \end{array}</math> | ||
|} | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | ! Arithmetic: | ||
| + | |- | ||
| + | |Now we simplify the numerator: | ||
| + | |- style = "text-align:center;" | ||
| + | | | ||
| + | <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{-6}{(3(x + h) + 1)(3x + 1))} | ||
| + | \end{array}</math> | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | ! Final Answer: | ||
| + | |- | ||
| + | |<math>\frac{-6}{(3(x + h) + 1)(3x + 1))}</math> | ||
| + | |} | ||
| + | [[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))}} |