Difference between revisions of "8A F11 Q2"
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|How would you find f(5) if f(x) = 2x + 1 instead? | |How would you find f(5) if f(x) = 2x + 1 instead? | ||
|- | |- | ||
| − | |Answer: we replace every occurrence of x with a 5. So f(5) = 2(5) + 1 = 11 | + | |Answer: we replace every occurrence of x with a 5. So |
| + | |- style = "text-align:center" | ||
| + | | | ||
| + | <math>\begin{array}{rcl} | ||
| + | f(5) &= &2(5) + 1\\ | ||
| + | & =& 11 | ||
| + | \end{array}</math> | ||
|} | |} | ||
| Line 18: | Line 24: | ||
! Step 1: | ! Step 1: | ||
|- | |- | ||
| − | |Replace any occurrence of x by 5, so <math>f(5) = \log_3(5 + 3) - 1 = \log_3(8) - 1</math> | + | |Replace any occurrence of x by 5, so |
| + | |- style = "text-align:center" | ||
| + | | | ||
| + | <math>\begin{array}{rcl} | ||
| + | f(5) &=& \log_3(5 + 3) - 1 \\ | ||
| + | &=& \log_3(8) - 1 \end{array}</math> | ||
|} | |} | ||
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|<math> \log_3(8) - 1</math> | |<math> \log_3(8) - 1</math> | ||
|} | |} | ||
| + | |||
| + | [[8AF11Final|<u>'''Return to Sample Exam</u>''']] | ||
Latest revision as of 15:22, 6 April 2015
Question: Find f(5) for f(x) given in problem 1.
Note: The function f(x) from problem 1 is:
| Foundations |
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| How would you find f(5) if f(x) = 2x + 1 instead? |
| Answer: we replace every occurrence of x with a 5. So |
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Solution:
| Step 1: |
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| Replace any occurrence of x by 5, so |
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| Final Answer: |
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