Difference between revisions of "8A F11 Q2"

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: $f(x)=\log _{3}(x+3)-1$

Foundations
How would you find f(5) if f(x) = 2x + 1 instead?
Answer: we replace every occurrence of x with a 5. So
${\begin{array}{rcl}f(5)&=&2(5)+1\\&=&11\end{array}}$

Solution:

Step 1:
Replace any occurrence of x by 5, so
${\begin{array}{rcl}f(5)&=&\log _{3}(5+3)-1\\&=&\log _{3}(8)-1\end{array}}$

Final Answer:
$\log _{3}(8)-1$

@@ Line 8: / Line 8: @@
 |-
 |How would you find f(5) if f(x) = 2x + 1 instead?
+|-
+|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 16: / Line 24: @@
 ! 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>
+|}
+{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
+! Final Answer:
+|-
+|<math> \log_3(8) - 1</math>
 |}
+[[8AF11Final|<u>'''Return to Sample Exam</u>''']]