Difference between revisions of "8A F11 Q12"
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(Created page with "'''Question: ''' Find and simplify the difference quotient <math>\frac{f(x+h)-f(x)}{h}</math> for f(x) = <math>\frac{2}{3x+1}</math> {| class="mw-collapsible mw-collapsed"...") |
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|Now we simplify the numerator: | |Now we simplify the numerator: | ||
|- style = "text-align:center;" | |- style = "text-align:center;" | ||
− | |<math>\frac{f(x + h) - f(x)}{h} = \left(\frac{2}{3(x + h) + 1} - \frac{2}{3x + 1}\right) \div h | + | | |
− | + | <math>\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}</math> | ||
|} | |} |
Revision as of 15:41, 6 April 2015
Question: Find and simplify the difference quotient for f(x) =
Foundations |
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1) f(x + h) = ? |
2) How do you eliminate the 'h' in the denominator? |
Answer: |
1) Since the difference quotient is a difference of fractions divided by h. |
2) The numerator is so the first step is to simplify this expression. This then allows us to eliminate the 'h' in the denominator. |
Solution:
Step 1: |
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The difference quotient that we want to simplify is |
Step 2: |
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Now we simplify the numerator: |
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